How can I calculate $intfracx-2-x^2+2x-5dx$?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I'm completely stuck on solving this indefinite integral:
$$intfracx-2-x^2+2x-5dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-intfracxx^2-2x+5dx -intfrac2(x-1)^2 + 4dx$$
The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.
integration indefinite-integrals partial-fractions
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
asked Jan 13 at 22:12
ArcturusArcturus
1117
$endgroup$
add a comment |
$begingroup$
I'm completely stuck on solving this indefinite integral:
$$intfracx-2-x^2+2x-5dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-intfracxx^2-2x+5dx -intfrac2(x-1)^2 + 4dx$$
The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.
integration indefinite-integrals partial-fractions
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
asked Jan 13 at 22:12
ArcturusArcturus
1117
$endgroup$
$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
Jan 13 at 22:17
$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
Jan 13 at 22:17
$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfracxdu(2x-2)u$?
$endgroup$
– Arcturus
Jan 13 at 22:27
$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
Jan 13 at 22:30
add a comment |
$begingroup$
I'm completely stuck on solving this indefinite integral:
$$intfracx-2-x^2+2x-5dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-intfracxx^2-2x+5dx -intfrac2(x-1)^2 + 4dx$$
The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.
integration indefinite-integrals partial-fractions
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
asked Jan 13 at 22:12
ArcturusArcturus
1117
$endgroup$
I'm completely stuck on solving this indefinite integral:
$$intfracx-2-x^2+2x-5dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-intfracxx^2-2x+5dx -intfrac2(x-1)^2 + 4dx$$
The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.
integration indefinite-integrals partial-fractions
integration indefinite-integrals partial-fractions
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
asked Jan 13 at 22:12
ArcturusArcturus
1117
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
asked Jan 13 at 22:12
ArcturusArcturus
1117
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
edited Jan 14 at 15:41
Asaf Karagila♦
303k32429761
303k32429761
asked Jan 13 at 22:12
ArcturusArcturus
1117
asked Jan 13 at 22:12
ArcturusArcturus
1117
asked Jan 13 at 22:12
ArcturusArcturus
1117
1117
$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
Jan 13 at 22:17
$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
Jan 13 at 22:17
$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfracxdu(2x-2)u$?
$endgroup$
– Arcturus
Jan 13 at 22:27
$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
Jan 13 at 22:30
add a comment |
$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
Jan 13 at 22:17
$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
Jan 13 at 22:17
$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfracxdu(2x-2)u$?
$endgroup$
– Arcturus
Jan 13 at 22:27
$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
Jan 13 at 22:30
$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
Jan 13 at 22:17
$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
Jan 13 at 22:17
$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
Jan 13 at 22:17
$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
Jan 13 at 22:17
$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfracxdu(2x-2)u$?
$endgroup$
– Arcturus
Jan 13 at 22:27
$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfracxdu(2x-2)u$?
$endgroup$
– Arcturus
Jan 13 at 22:27
$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
Jan 13 at 22:30
$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
Jan 13 at 22:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
beginalignintfracx-2-x^2+2x-5dx &= - int fracx-2x^2 - 2x + 5dx
\&= -frac 12 int frac2x-4x^2 - 2x + 5dx. endalign The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
beginalign
intfracx-2-x^2+2x-5dx = -frac 12 left(int frac2x-2x^2-2x+5dx - int frac2x^2-2x+5dxright).
endalign
The second integral you can solve, can you solve the first?
answered Jan 13 at 22:30
E-muE-mu
786417
$endgroup$
2
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
add a comment |
$begingroup$
Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
beginalign*I &= -int fracx-2(x-1)^2+4,dx\
&phantom^u=x-1_du=dx\
&= int -fracu-1u^2+4,du = intfrac-uu^2+4,du+intfrac1u^2+4,duendalign*
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.
answered Jan 13 at 23:06
jmerryjmerry
5,987718
$endgroup$
add a comment |
$begingroup$
Since $x^2-2x+5=(x-1)^2+2^2$, the substitution to do is $2u=x-1$, so the integral becomes
$$
intfrac2-xx^2-2x+5,dx=
intfrac1-2u4u^2+42,du=frac12intfrac1-2uu^2+1,du
$$
No need to guess, now.
answered Jan 14 at 8:22
egregegreg
181k1485202
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072589%2fhow-can-i-calculate-int-fracx-2-x22x-5dx%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
beginalignintfracx-2-x^2+2x-5dx &= - int fracx-2x^2 - 2x + 5dx
\&= -frac 12 int frac2x-4x^2 - 2x + 5dx. endalign The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
beginalign
intfracx-2-x^2+2x-5dx = -frac 12 left(int frac2x-2x^2-2x+5dx - int frac2x^2-2x+5dxright).
endalign
The second integral you can solve, can you solve the first?
answered Jan 13 at 22:30
E-muE-mu
786417
$endgroup$
2
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
add a comment |
$begingroup$
The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
beginalignintfracx-2-x^2+2x-5dx &= - int fracx-2x^2 - 2x + 5dx
\&= -frac 12 int frac2x-4x^2 - 2x + 5dx. endalign The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
beginalign
intfracx-2-x^2+2x-5dx = -frac 12 left(int frac2x-2x^2-2x+5dx - int frac2x^2-2x+5dxright).
endalign
The second integral you can solve, can you solve the first?
answered Jan 13 at 22:30
E-muE-mu
786417
$endgroup$
2
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
add a comment |
$begingroup$
The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
beginalignintfracx-2-x^2+2x-5dx &= - int fracx-2x^2 - 2x + 5dx
\&= -frac 12 int frac2x-4x^2 - 2x + 5dx. endalign The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
beginalign
intfracx-2-x^2+2x-5dx = -frac 12 left(int frac2x-2x^2-2x+5dx - int frac2x^2-2x+5dxright).
endalign
The second integral you can solve, can you solve the first?
answered Jan 13 at 22:30
E-muE-mu
786417
$endgroup$
The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
beginalignintfracx-2-x^2+2x-5dx &= - int fracx-2x^2 - 2x + 5dx
\&= -frac 12 int frac2x-4x^2 - 2x + 5dx. endalign The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
beginalign
intfracx-2-x^2+2x-5dx = -frac 12 left(int frac2x-2x^2-2x+5dx - int frac2x^2-2x+5dxright).
endalign
The second integral you can solve, can you solve the first?
answered Jan 13 at 22:30
E-muE-mu
786417
edited Jan 13 at 22:38
answered Jan 13 at 22:30
E-muE-mu
786417
answered Jan 13 at 22:30
E-muE-mu
786417
answered Jan 13 at 22:30
E-muE-mu
786417
786417
2
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
add a comment |
2
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
2
2
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
$begingroup$
Thank you for a concise and very easy to understand answer. It's an ultimately simple trick of manipulating the expression to your favor, but the hard part is thinking of it. Well, hard at this late hour for me, at least :D Cheers mate. This will make any following similar problems that much easier to solve.
$endgroup$
– Arcturus
Jan 13 at 22:58
add a comment |
$begingroup$
Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
beginalign*I &= -int fracx-2(x-1)^2+4,dx\
&phantom^u=x-1_du=dx\
&= int -fracu-1u^2+4,du = intfrac-uu^2+4,du+intfrac1u^2+4,duendalign*
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.
answered Jan 13 at 23:06
jmerryjmerry
5,987718
$endgroup$
add a comment |
$begingroup$
Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
beginalign*I &= -int fracx-2(x-1)^2+4,dx\
&phantom^u=x-1_du=dx\
&= int -fracu-1u^2+4,du = intfrac-uu^2+4,du+intfrac1u^2+4,duendalign*
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.
answered Jan 13 at 23:06
jmerryjmerry
5,987718
$endgroup$
add a comment |
$begingroup$
Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
beginalign*I &= -int fracx-2(x-1)^2+4,dx\
&phantom^u=x-1_du=dx\
&= int -fracu-1u^2+4,du = intfrac-uu^2+4,du+intfrac1u^2+4,duendalign*
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.
answered Jan 13 at 23:06
jmerryjmerry
5,987718
$endgroup$
Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
beginalign*I &= -int fracx-2(x-1)^2+4,dx\
&phantom^u=x-1_du=dx\
&= int -fracu-1u^2+4,du = intfrac-uu^2+4,du+intfrac1u^2+4,duendalign*
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.
answered Jan 13 at 23:06
jmerryjmerry
5,987718
answered Jan 13 at 23:06
jmerryjmerry
5,987718
answered Jan 13 at 23:06
jmerryjmerry
5,987718
answered Jan 13 at 23:06
jmerryjmerry
5,987718
5,987718
add a comment |
add a comment |
$begingroup$
Since $x^2-2x+5=(x-1)^2+2^2$, the substitution to do is $2u=x-1$, so the integral becomes
$$
intfrac2-xx^2-2x+5,dx=
intfrac1-2u4u^2+42,du=frac12intfrac1-2uu^2+1,du
$$
No need to guess, now.
answered Jan 14 at 8:22
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
Since $x^2-2x+5=(x-1)^2+2^2$, the substitution to do is $2u=x-1$, so the integral becomes
$$
intfrac2-xx^2-2x+5,dx=
intfrac1-2u4u^2+42,du=frac12intfrac1-2uu^2+1,du
$$
No need to guess, now.
answered Jan 14 at 8:22
egregegreg
181k1485202
$endgroup$
add a comment |
$begingroup$
Since $x^2-2x+5=(x-1)^2+2^2$, the substitution to do is $2u=x-1$, so the integral becomes
$$
intfrac2-xx^2-2x+5,dx=
intfrac1-2u4u^2+42,du=frac12intfrac1-2uu^2+1,du
$$
No need to guess, now.
answered Jan 14 at 8:22
egregegreg
181k1485202
$endgroup$
Since $x^2-2x+5=(x-1)^2+2^2$, the substitution to do is $2u=x-1$, so the integral becomes
$$
intfrac2-xx^2-2x+5,dx=
intfrac1-2u4u^2+42,du=frac12intfrac1-2uu^2+1,du
$$
No need to guess, now.
answered Jan 14 at 8:22
egregegreg
181k1485202
answered Jan 14 at 8:22
egregegreg
181k1485202
answered Jan 14 at 8:22
egregegreg
181k1485202
answered Jan 14 at 8:22
egregegreg
181k1485202
181k1485202
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072589%2fhow-can-i-calculate-int-fracx-2-x22x-5dx%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Hint: look at the derivative of $ln (x^2-2x+5)$ to see how it helps.
$endgroup$
– J.G.
Jan 13 at 22:17
$begingroup$
Let $u$ be the denominator and notice how $du$ nicely falls out.
$endgroup$
– John Douma
Jan 13 at 22:17
$begingroup$
@JohnDouma That's what I tried, but if $u$ is the denominator, aren't I left with $intfracxdu(2x-2)u$?
$endgroup$
– Arcturus
Jan 13 at 22:27
$begingroup$
You could refer to an online integral calculator. integral-calculator.com It has a complete solution. It is a rather long solution, so I feel kind of lazy to type them out.
$endgroup$
– Larry
Jan 13 at 22:30