## Jonask 2 years ago nth integral

Let $I_n=\huge\int \frac{x^n}{\sqrt{x^2+a}}$ $n \in N,a>0,and \space I_n=(a,\infty)$

Find $I_0,I_1 ,I_n$

I did the first two but three i dont understand

actually it says show that $I_n=\frac{1}{n}x^{n-1}\sqrt{x^2+a}-\frac{n-1}{n}aI_{n-2}$

$I_0=\ln(x+\sqrt{x^2+a})-\ln\sqrt{a}+c,I_1=\frac{1}{2}\sqrt{x^2+a}+c$

6. hartnn

tried integrating by parts ?

yes$f=x^n,f'=nx^{n-1}$ $g'=\frac{1}{\sqrt{x^2+a}}.g=I_0$ $I_n=x^n(I_0)-\int I_0nx^{n-1}$

$I_0$ is known above

$I_n=x^n\ln(x+\sqrt{x^2+a})-n\int x^{n-1}\ln(x+\sqrt{x^2+a})$

using parts twice $\int(x^{n-1})\ln(x+\sqrt{x^2+a)}$ $f=x^n-1,f'=(n-1)x^{n-2}$ $g'=\ln(x+\sqrt{x^2+a)}$ not sure wat to use for g

sry $f=x^{n-1}$

12. hartnn

let me try lil bit different, $$I_n=\huge\int \frac{x^n}{\sqrt{x^2+a}}dx=\huge\int \frac{x.x^{n-1}}{\sqrt{x^2+a}}dx$$ now let f(x) = x^{n-1} , g'(x) =x/(...)

13. hartnn

just a thought, donno whether it'll be useful

thanks it works very well,i tried it,just too lazy to post

15. hartnn

oh, it works...good to hear :)