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Help- Integral by Parts

Chipset3600

Member
Feb 14, 2012
79
Hello MHB, i'm trying to solve this integral:[TEX]\int x^2arctan(x)dx[/TEX] by parts :
here is a litle bit of my resolution: http://i.imgur.com/dZk8M.jpg

when I tried to solve the integral I named B fell into a sort of lool lool.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Hello MHB, i'm trying to solve this integral:[TEX]\int x^2arctan(x)dx[/TEX] by parts :
here is a litle bit of my resolution: http://i.imgur.com/dZk8M.jpg

when I tried to solve the integral I named B fell into a sort of lool lool.
All looks good so far. Let's focus on this guy, though:

\[\int\frac{x^3}{1+x^2}\,dx\]

I would first recommend you reduce the integrand using long division. Doing this correctly should yield

\[\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}\]

So, we see that

\[\int\frac{x^3}{1+x^2}\,dx=\int\left(x-\frac{x}{1+x^2}\right)\,dx\]

This should be a straightforward integration; no IBP is needed at this point.

Combine this result with everything else you have and you'll get your answer.

I hope this makes sense!
 

Chipset3600

Member
Feb 14, 2012
79
All looks good so far. Let's focus on this guy, though:

\[\int\frac{x^3}{1+x^2}\,dx\]

I would first recommend you reduce the integrand using long division. Doing this correctly should yield

\[\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}\]

So, we see that

\[\int\frac{x^3}{1+x^2}\,dx=\int\left(x-\frac{x}{1+x^2}\right)\,dx\]

This should be a straightforward integration; no IBP is needed at this point.

Combine this result with everything else you have and you'll get your answer.

I hope this makes sense!
Thanks, it was much simpler, I tried to do polynomial division, but not crossed my mind to separate fractions by.
Thank you
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I have tried another method: use the substitution $u = 1+x^2$. Therefore $du = 2x \, dx$ and $x^2 = u-1$. The integral changes shape into

$$\int \frac{x^3}{1+x^2} \, dx = \frac{1}{2} \int (u-1) \, du.$$

This yields

$$\frac{1}{2} \int (u-1) \, du = \frac{u^2}{4} + \frac{u}{2} + C = \frac{(x^2 +1)^2}{4} + \frac{(x^2 +1)}{2} + C.$$

However, it is a different answer. What is wrong?
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
I have tried another method: use the substitution $u = 1+x^2$. Therefore $du = 2x \, dx$ and $x^2 = u-1$. The integral changes shape into

$$\int \frac{x^3}{1+x^2} \, dx = \frac{1}{2} \int (u-1) \, du.$$

This yields

$$\frac{1}{2} \int (u-1) \, du = \frac{u^2}{4} + \frac{u}{2} + C = \frac{(x^2 +1)^2}{4} + \frac{(x^2 +1)}{2} + C.$$

However, it is a different answer. What is wrong?
Your integrand after making the substitution isn't correct. You should have

\[\int \frac{x^3}{1+x^2}\,dx\xrightarrow{u=x^2+1}{}\int \frac{u-1}{2u}\,du.\]
 
Last edited:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Good point, I had forgotten that part. Now the answer is complete (and easier). Thanks Chris! (Clapping)