Question

nth integral

Answer 1

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

Answer 2

Find \[I_0,I_1 ,I_n\]

Answer 3

I did the first two but three i dont understand

Answer 4

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

Answer 5

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

Answer 6

tried integrating by parts ?

Answer 7

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

Answer 8

\[I_0\] is known above

Answer 9

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

Answer 10

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

Answer 11

sry \[f=x^{n-1}\]

Answer 12

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/(...)

Answer 13

just a thought, donno whether it'll be useful

Answer 14

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

Answer 15

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