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anonymous
 3 years ago
nth integral
anonymous
 3 years ago
nth integral

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Find \[I_0,I_1 ,I_n\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did the first two but three i dont understand

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually it says show that \[I_n=\frac{1}{n}x^{n1}\sqrt{x^2+a}\frac{n1}{n}aI_{n2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[I_0=\ln(x+\sqrt{x^2+a})\ln\sqrt{a}+c,I_1=\frac{1}{2}\sqrt{x^2+a}+c\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes\[ f=x^n,f'=nx^{n1}\] \[g'=\frac{1}{\sqrt{x^2+a}}.g=I_0\] \[I_n=x^n(I_0)\int I_0nx^{n1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[I_0\] is known above

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[I_n=x^n\ln(x+\sqrt{x^2+a})n\int x^{n1}\ln(x+\sqrt{x^2+a})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0using 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0just a thought, donno whether it'll be useful

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks it works very well,i tried it,just too lazy to post

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