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

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

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

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

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

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

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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.2using 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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0let 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/(...)

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

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

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