Question

Find the generalized form of the integral: (X^n)/(1+x^2)

Answer 1

@wio

Answer 2

@shrutipande9

Answer 3

for any positive integer n,

Answer 4

@ganeshie8

Answer 5

use trig substitution as x= tan theta.
so
\[\int\limits_{?}^{?}\frac{ x^n }{ 1+x^2 }dx = \int\limits_{?}^{?} \tan^n \theta d \theta=\int\limits_{?}^{?}(1+\sec^2)\tan ^{n-2}\theta d \theta\]
I you can continue to get the final form.

Answer 6

Please in order to get your answer, it is necessary to evaluate your integral for n=0,1
and then we have to consider the cases n odd and n even.
I got your solution, nevertheless I can not post it since the Code-of Conduct

Answer 7

excuse me @Michele_Laino what is Code-of Conduct?

Answer 8

please see the link below:
http://openstudy.com/code-of-conduct
@Catch.me

Answer 9

@michele_laino can you solve it for me?

Answer 10

\[\int\limits_{0}^{1}\frac{ 1 }{1+x^2} dx =\tan^{-1} x \]

Answer 11

\[\int\limits_{0}^{1}\frac{ x }{ 1+x^2 } dx= \frac{ 1 }{ 2 }\ln (1+x^2)\]

Answer 12

\[\int\limits_{0}^{1}\frac{ x^2 }{ 1+x^2} dx = x-\tan^{-1} x\]

Answer 13

\[\int\limits_{0}^{1} \frac{ x^3 }{ 1+x^2 } dx = 1/2x^2 -1/2\ln(x^2+1)\]

Answer 14

For \(n\ge3\), you have the recursive formula
\[I_n=\int\frac{x^n}{1+x^2}\,dx=\int\left(x^{n-2}-\frac{x^{n-2}}{1+x^2}\right)\,dx=\int x^{n-2}\,dx-I_{n-2}\]
Judging by the alternating pattern between \(\arctan\) and \(\ln\) terms above, I would expect a different result for even and odd \(n\). Try breaking up into two cases, with \(k=1,2,3,...\) (such that \(2k+1=3\) and \(2k=4\), and so on):
\[I_{2k}=\int\frac{x^{2k}}{1+x^2}\,dx\quad\text{and}\quad I_{2k+1}=\int\frac{x^{2k+1}}{1+x^2}\,dx\]
For the latter, a substitution of \(u=1+x^2\) gives \(\dfrac{du}{2}=x\,dx\), and so
\[I_{2k+1}=\frac{1}{2}\int \frac{(u-1)^k}{u}\,du\]
This can be expanded using the binomial theorem and integrated term by term.

For the even integral, setting \(x=\tan t\) gives \(dx=\sec^2t\,dt\), hence
\[\begin{align*}I_{2k}&=\int\frac{\tan^{2k}t\sec^2t}{1+\tan^2t}\,dt\\\\&=\int\tan^{2k}t\,dt\\\\&=\int(\tan^{2k-2}t\sec^2t-\tan^{2k-2}t)\,dt\\\\
&=\frac{\tan^{2k-1}t}{2k-1}-\int\tan^{2k-2}t\,dt\\\\
I_{2k}&=\frac{\tan^{2k-1}t}{2k-1}-I_{2k-2}\end{align*}\]