Question

prove that
\[\int \frac{dx}{(x^2+1)^{n+1}}=\frac{x}{2n(x^2+1)^n}+(1-\frac{1}{2n})\int \frac{dx}{(x^2+1)^n}\]

Answer 1

proved by @nnesha

Answer 2

proved by @AdamK

Answer 3

this supposed to be by parts i tried dozen of things hahha

Answer 4

@Nnesha you want to try haha

Answer 5

@freckles
another problem here lol

Answer 6

hehe just give it a try you never know lol

Answer 7

hahah :)

Answer 8

\[\int\limits_{}^{}\frac{x}{(x^2+1)^{n+1}} dx \\ \frac{1}{2} \int\limits \frac{ du}{u^{n+1}}=\frac{1}{2}\frac{u^{-n-1+1}}{-n-1+1}+C= \frac{-1}{2n u^n}+C=\frac{-1}{2n(x^2+1)^{n}} +C \\ (\frac{1}{x})'=\frac{-1}{x^2} \\ \int\limits \frac{1}{x} \cdot \frac{x}{(x^2+1)^{n+1}} dx= \\ \frac{1}{x} \cdot \frac{-1}{2n (x^2+1)^n}- \int\limits (\frac{-1}{x^2}) \frac{-1}{2n(x^2+1)^n} dx \\ \]
hmmm but this doesn't looks like it helps

Answer 9

we don't have x on the top for that integrand
it is just \[\int \frac{dx}{(x^2+1)^{n+1}}\]

Answer 10

@perl some help :)

Answer 11

Yeah I know I multiplied by x/x

Answer 12

see bottom 2 lines

Answer 13

oh you got me for the first line haha
i thought you did something different
actually i tried that me and some friends but we didn't really got an interesting thing
that look like what we needed

Answer 14

I don't know I didn't get an x in the numerator :(

Answer 15

for the first term thing

Answer 16

that x and x^2 make a problem for us

Answer 17

i tried with u=x^2+1 then use by parts later

Answer 18

my integral
\[\int \frac{du}{\sqrt{u-1} u^{n+1}}\]

Answer 19

of times 1/2

Answer 20

forgot 1/2 there

Answer 21

The proof is attached. @xapproachesinfinity @freckles @ganeshie8

Answer 22

thanks a lot @eliassaab

Answer 23

YW