# recursive integration (1 Viewer)

#### 5uckerberg

##### Well-Known Member
Let's do this properly.
$\bg_white I_{n}=\int_{0}^{1}\frac{x^{2}}{\left(1+x^{2}\right)^{n}}dx=\int_{0}^{1}x^{2}\left(1+x^{2}\right)^{-n}dx$
After a painstaking battle, I have found that this question can easily get overcomplicated by the student.
The first step integrate $\bg_white x^{2}$
By doing so we will have $\bg_white \left[\frac{x^{3}}{3}\left(1+x^{2}\right)^{-n}\right]_{0}^{1}$
Sub in x=1 and we will have $\bg_white \frac{2^{-n}}{3}$ and x=0 will give us 0.
Fear not have faith in yourself.

Now we will be differentiating $\bg_white \left(1+x^{2}\right)^{-n}$
The chain rule is your friend $\bg_white \frac{d}{dx}\left(1+x^{2}\right)^{-n}=-2nx\left(1+x^{2}\right)^{-n-1}=-\frac{2nx}{\left(1+x^{2}\right)^{n+1}}$
Multiply everything by $\bg_white -\frac{x^{3}}{3}$ and we will have $\bg_white \frac{2nx^{4}}{3\left(1+x^{2}\right)^{n+1}}$. In conclusion we are workin g with
$\bg_white I_{n}=\int_{0}^{1}\frac{x^{2}}{\left(1+x^{2}\right)^{n}}dx=\int_{0}^{1}x^{2}\left(1+x^{2}\right)^{-n}dx=\frac{2^{-n}}{3}+\int_{0}^{1}\frac{2nx^{4}}{3\left(1+x^{2}\right)^{n+1}}dx$
Simply put, we have $\bg_white I_{n}=\frac{2^{-n}}{3}+\int_{0}^{1}\frac{2nx^{4}}{3\left(1+x^{2}\right)^{n+1}}dx$.
Now, multiply by 3
$\bg_white 3I_{n}=2^{-n}+\int_{0}^{1}\frac{2nx^{4}}{\left(1+x^{2}\right)^{n+1}}dx$
$\bg_white 0=2^{-n}-3I_{n}+\int_{0}^{1}\frac{2nx^{4}}{\left(1+x^{2}\right)^{n+1}}dx$

Now what?

Keep an eye on this term $\bg_white \int_{0}^{1}\frac{2nx^{4}}{\left(1+x^{2}\right)^{n+1}}dx$
What can you do to get something like $\bg_white \frac{x^{2}}{\left(1+x^{2}\right)^{n}}$?

If you can solve that step you will obtain a huge morale boost.

$\bg_white \int_{0}^{1}\frac{2nx^{4}}{\left(1+x^{2}\right)^{n+1}}dx=\frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}\times{\left(\frac{x^{2}}{1+x^{2}}\right)$
$\bg_white \frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}\times{\left(\frac{x^{2}}{1+x^{2}}\right)=\frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}\times\left(1-\frac{1}{1+x^{2}}\right)$
Let's finish this the final stretch.
$\bg_white \frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}\times\left(1-\frac{1}{1+x^{2}}\right)=\frac{2nx^{2}}{\left(1+x^{2}\right)^{n}}-\frac{2nx^{2}}{\left(1+x^{2}\right)^{n+1}}=2nI_{n}-2nI_{n+1}$
Combining it all together
$\bg_white 0=2^{-n}-3I_{n}+2nI_{n}-2nI_{n+1}$.
Rearranging we will then have
$\bg_white 2nI_{n+1}=2^{-n}+2nI_{n}-3I_{n}=2^{-n}+\left(2n-3\right)I_{n}$