Reduction Integral

Real Analysis
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jacks
Posts: 102
Joined: Thu Nov 12, 2015 5:26 pm
Location: Himachal Pradesh (INDIA)

Reduction Integral

#1

Post by jacks »

If \(\displaystyle I_{n} = \int\frac{x^n}{(ax^2+bx+c)^{\frac{1}{2}}}dx\) and \(\displaystyle n\in \mathbb{N}\). Then value of \(\displaystyle I_{n+1}\) in terms of \(\displaystyle I_{n}\) and \(\displaystyle I_{n-1}\)
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Reduction Integral

#2

Post by Papapetros Vaggelis »

If \(\displaystyle{n\in\mathbb{N}}\) and \(\displaystyle{a\neq 0}\) , then :

\(\displaystyle{\begin{aligned} I_{n+1}&=\int \dfrac{x^{n+1}}{\sqrt{a\,x^2+b\,x+c}}\,\mathrm{d}x\\&=\int \dfrac{\left(2\,a\,x+b\right)\,x^{n}-b\,x^{n}}{2\,a\,\sqrt{a\,x^2+b\,x+c}}\,\mathrm{d}x\\&=\dfrac{1}{2\,a}\,\int \left(\dfrac{\left(2\,a\,x+b\right)\,x^{n}}{\sqrt{a\,x^2+b\,x+c}}-b\,\dfrac{x^{n}}{\sqrt{a\,x^2+b\,x+c}}\right)\,\mathrm{d}x\\&=\dfrac{1}{2\,a}\,\left[2\,x^{n}\,\sqrt{a\,x^2+b\,x+c}-2\,n\,\int \sqrt{a\,x^2+b\,x+c}\,x^{n-1}\,\mathrm{d}x\right]-\dfrac{b}{2\,a}\,I_{n}\\&=\dfrac{x^{n}\,\sqrt{a\,x^2+b\,x+c}}{a}-\dfrac{n}{a}\,\int \dfrac{x^{n-1}\,\left(a\,x^2+b\,x+c\right)}{\sqrt{a\,x^2+b\,x+c}}\,\mathrm{d}x-\dfrac{b}{2\,a}\,I_{n}\\&=\dfrac{x^{n}\,\sqrt{a\,x^2+b\,x+c}}{a}-\dfrac{n}{a}\,\int \dfrac{a\,x^{n+1}+b\,x^{n}+c\,x^{n-1}}{\sqrt{a\,x^2+b\,x+c}}\,\mathrm{d}x-\dfrac{b}{2\,a}\,I_{n}\\&=\dfrac{x^{n}\,\sqrt{a\,x^2+b\,x+c}}{a}-n\,I_{n+1}-\dfrac{b\,n}{a}\,I_{n}-\dfrac{c\,n}{a}\,I_{n-1}\end{aligned}}\)

so :

\(\displaystyle{\left(n+1\right)\,I_{n+1}=\dfrac{x^{n}\,\sqrt{a\,x^2+b\,x+c}}{a}-\dfrac{b\,\left(2\,n+1\right)}{2\,a}\,I_{n}-\dfrac{c\,n}{a}\,I_{n-1}}\)

or

\(\displaystyle{I_{n+1}=\dfrac{x^{n}\,\sqrt{a\,x^2+b\,x+c}}{a\,\left(n+1\right)}-\dfrac{b\,\left(2\,n+1\right)}{2\,a\,\left(n+1\right)}\,I_{n}-\dfrac{c\,n}{a\,\left(n+1\right)}\,I_{n-1}}\)
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