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JonaskBest ResponseYou've already chosen the best response.2
Let \[I_n=\huge\int \frac{x^n}{\sqrt{x^2+a}}\] \[n \in N,a>0,and \space I_n=(a,\infty)\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
I did the first two but three i dont understand
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
actually it says show that \[I_n=\frac{1}{n}x^{n1}\sqrt{x^2+a}\frac{n1}{n}aI_{n2}\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
\[I_0=\ln(x+\sqrt{x^2+a})\ln\sqrt{a}+c,I_1=\frac{1}{2}\sqrt{x^2+a}+c\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
tried integrating by parts ?
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
yes\[ f=x^n,f'=nx^{n1}\] \[g'=\frac{1}{\sqrt{x^2+a}}.g=I_0\] \[I_n=x^n(I_0)\int I_0nx^{n1}\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
\[I_n=x^n\ln(x+\sqrt{x^2+a})n\int x^{n1}\ln(x+\sqrt{x^2+a})\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
using parts twice \[\int(x^{n1})\ln(x+\sqrt{x^2+a)}\] \[f=x^n1,f'=(n1)x^{n2}\] \[g'=\ln(x+\sqrt{x^2+a)}\] not sure wat to use for g
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
let me try lil bit different, \(I_n=\huge\int \frac{x^n}{\sqrt{x^2+a}}dx=\huge\int \frac{x.x^{n1}}{\sqrt{x^2+a}}dx\) now let f(x) = x^{n1} , g'(x) =x/(...)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
just a thought, donno whether it'll be useful
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
thanks it works very well,i tried it,just too lazy to post
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
oh, it works...good to hear :)
 one year ago
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