if I(n)=\int \frac{x^{n}}{(ax^{2}+bx+c)^{1/2}} dx
n belongs to N
find I(n+1) in terms of I(n) and I(n-1)
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1 Answers
11x^n=\frac{2ax+b-b}{2a}x^{n-1}
Now applying by-parts we have
I_n=\frac{1}{2a}\int {\frac{2ax+b}{\sqrt{ax^2+bx+c}}dx -\frac{b}{2a}\int {\frac{x^{n-1}}{\sqrt{ax^2+bx+c}}dx
Finally we apply by-parts to have I_n=\frac{x^{n-1}}{a}\sqrt{ax^2+bx+c}-\frac{n-1}{a}\left\{aI_n+bI_{n-1}+cI_{n-2} \right\}-\frac{b}{2a}I_{n-1}
On slight simplification we can thus get our reqd answer.....
