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Question Number 213376 by RoseAli last updated on 03/Nov/24

$$\int \frac{dx}{\sqrt{{(4x^2+1)}^3}}$$

Commented by Frix last updated on 03/Nov/24

$$\mathrm{Sometimes}\mathrm{just}\mathrm{use}\mathrm{your}\mathrm{brain}\mathrm{\&}\mathrm{experience}$$$$\frac{d}{dx}\left[\frac{g\left(x\right)}{\sqrt{4x^2+1}}\right]=\frac{g^{\prime }\left(x\right)(4x^2+1)-4g\left(x\right)}{{(4x^2+1)}^{\frac{3}{2}}}$$$$g^{\prime }\left(x\right)(4x^2+1)-4g\left(x\right)=1\Rightarrow g\left(x\right)=x$$$$\int \frac{dx}{\sqrt{{(4x^2+1)}^3}}=\frac{x}{\sqrt{4x^2+1}}+C$$

Answered by Frix last updated on 03/Nov/24

$$\int \frac{dx}{{(4x^2+1)}^{\frac{3}{2}}}\stackrel{[t={\tan }^{-1}2x]}{=}\frac{1}{2}\int \cos tdt=\frac{1}{2}\sin t=$$$$=\frac{x}{\sqrt{4x^2+1}}+C$$

Answered by Frix last updated on 03/Nov/24

$$\int \frac{dx}{{(4x^2+1)}^{\frac{3}{2}}}\stackrel{[t=2x+\sqrt{4x^2+1}]}{=}\int \frac{2t}{{(t^2+1)}^2}dt=$$$$=-\frac{1}{t^2+1}=\frac{x}{\sqrt{4x^2+1}}+C$$

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