djeljenike

2020-11-01

Evaluate the integrals. $\int \frac{{\mathrm{tan}}^{-1}x}{{x}^{2}}dx$

pierretteA

$\text{Given}$
$\int \frac{{\mathrm{tan}}^{-1}x}{{x}^{2}}dx$
$\text{Apply integration by parts:}\int f\cdot {g}^{\prime }=fg-\int {f}^{\prime }\cdot g$

$\int f\cdot {g}^{\prime }=-\frac{{\mathrm{tan}}^{-1}\left(x\right)}{x}-\int -\frac{1}{x\left({x}^{2}+1\right)}dx$
$\int -\frac{1}{x\left({x}^{2}+1\right)}dx=-\int \frac{1}{x\left({x}^{2}+1\right)}dx$
$\text{The integral solce by expanding the fraction by}\frac{1}{{x}^{3}}$
$\int \frac{1}{x\left({x}^{2}+1\right)}dx=\int \frac{1}{\left(\frac{1}{{x}^{2}}+1\right){x}^{3}}dx$

$\int \frac{1}{\left(\frac{1}{{x}^{2}}+1\right){x}^{3}}dx=\int \frac{1}{\left(u\right){x}^{3}}×-\frac{{x}^{3}}{2}du=-\frac{1}{2}\int \frac{1}{u}du$
$=-\frac{\mathrm{ln}\left(u\right)}{2}$

$=-\frac{\mathrm{ln}\left(\frac{1}{{x}^{2}}+1\right)}{2}$
$⇒-\int \frac{1}{x\left({x}^{2}+1\right)}dx=\frac{\mathrm{ln}\left(\frac{1}{{x}^{2}}+1\right)}{2}$
$⇒-\frac{{\mathrm{tan}}^{-1}\left(x\right)}{x}-\int -\frac{1}{x\left({x}^{2}+1\right)}dx=\frac{{\mathrm{tan}}^{-1}\left(x\right)}{x}-\frac{\mathrm{ln}\left(\frac{1}{{x}^{2}}+1\right)}{2}+C$
$\int \frac{{\mathrm{tan}}^{-1}x}{{x}^{2}}dx=\frac{{\mathrm{tan}}^{-1}\left(x\right)}{x}-\frac{\mathrm{ln}\left(\frac{1}{{x}^{2}}+1\right)}{2}+C$

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