MHB Efficient Integral by Parts Method | Solving x^2arctan(x)dx with Ease

[Chipset3600]

Hello MHB, I'm trying to solve this integral:\int x^2arctan(x)dx 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]

Hello MHB, I'm trying to solve this integral:\int x^2arctan(x)dx 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]

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]

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]

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

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

 

[Fantini]

Good point, I had forgotten that part. Now the answer is complete (and easier). Thanks Chris! (Clapping)