- Thread starter
- #1

#### Chipset3600

##### Member

- Feb 14, 2012

- 79

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.

- Thread starter Chipset3600
- Start date

- Thread starter
- #1

- Feb 14, 2012

- 79

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.

- Moderator
- #2

- Jan 26, 2012

- 995

All looks good so far. Let's focus on this guy, though:

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.

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

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!

- Thread starter
- #3

- Feb 14, 2012

- 79

Thanks, it was much simpler, I tried to do polynomial division, but not crossed my mind to separate fractions by.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!

Thank you

- Feb 29, 2012

- 342

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

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?

- Moderator
- #5

- Jan 26, 2012

- 995

Your integrand after making the substitution isn't correct. You should have

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

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?

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

Last edited:

- Feb 29, 2012

- 342

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