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Question Number 133183 by bounhome last updated on 19/Feb/21

$$\int \sqrt{{(x^2+1)}^3}dx=...?$$

Answered by physicstutes last updated on 19/Feb/21

$$\mathrm{let}x=\sinh \theta \Rightarrow dx=\cosh \theta d\theta$$$$\mathrm{now},\int \sqrt{{[{\left(\sinh \theta \right)}^2+1]}^3}\cosh \theta d\theta$$$$\Rightarrow \int \sqrt{{\left({\cosh }^2\theta \right)}^3}\cosh \theta d\theta =\int \sqrt{{\cosh }^6\theta }\cosh \theta d\theta =\int {\cosh }^4\theta d\theta$$$$\mathcal{I}=\int {\cosh }^4\theta d\theta =\frac{1}{32}(12\theta +8\sinh 2\theta +\sinh 4\theta )+A,A\in \mathbb{R}$$

Commented by Dwaipayan Shikari last updated on 19/Feb/21

$$\frac{12}{32}\theta +\frac{1}{2}sinh\theta cosh\theta +\frac{1}{8}sinh\theta cosh\theta (1-2{sinh}^2\theta )$$$$=\frac{3}{8}{sinh}^{-1}x+\frac{x\sqrt{1+x^2}}{2}+\frac{x\sqrt{1+x^2}}{8}(1-2x^2)+C$$$$=\frac{3}{8}log(x+\sqrt{1+x^2})+\frac{x\sqrt{1+x^2}}{8}(5-2x^2)+C$$

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