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Question Number 128334 by liberty last updated on 06/Jan/21

$$\mathrm{\Omega }=\int \frac{{\mathrm{x}}^2}{\sqrt{{(\mathrm{a}+{\mathrm{bx}}^2)}^5}}\mathrm{dx};\mathrm{where}:\mathrm{a};\mathrm{b}>0$$

Answered by bramlexs22 last updated on 06/Jan/21

$$\mathrm{\Omega }=\int \frac{x^2}{{(a+{bx}^2)}^{5/2}}dx$$ $$\mathrm{\Omega }=\int \frac{x^2}{{\left(x^2({ax}^{-2}+b)\right)}^{5/2}}dx$$ $$\mathrm{\Omega }=\int \frac{x^{-3}}{{({ax}^{-2}+b)}^{5/2}}dx$$ $$setting{ax}^{-2}+b=v$$ $$weget-2{ax}^{-3}dx=dv$$ $$then\mathrm{\Omega }=-\frac{1}{2a}\int \frac{dv}{v^{5/2}}$$ $$\mathrm{\Omega }=-\frac{1}{2a}\int v^{-5/2}dv=-\frac{1}{2a}.(-\frac{2}{3})v^{-3/2}+C$$ $$\mathrm{\Omega }=\frac{1}{3a}.\frac{1}{\sqrt{v^3}}+C=\frac{1}{3a\sqrt{{\left(\frac{a+{bx}^2}{x^2}\right)}^3}}+C$$ $$\mathrm{\Omega }=\frac{x^3}{3a\sqrt{{(a+{bx}^2)}^3}}+C$$

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