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Question Number 40870 by math khazana by abdo last updated on 28/Jul/18

$$fnd\int (1+\frac{1}{x^2})arctan(x-\frac{1}{x})dx.$$

Commented by maxmathsup by imad last updated on 29/Jul/18

$$changementx-\frac{1}{x}=tgive(1+\frac{1}{x^2})dx=dt\Rightarrow$$$$I=\int arctantdt=tarctant-\int \frac{t}{1+t^2}dt+c$$$$=tarctan\left(t\right)-\frac{1}{2}ln(1+t^2)+c$$$$=(x-\frac{1}{x})arctan(x-\frac{1}{x})-\frac{1}{2}ln(1+{(x-\frac{1}{x})}^2)+c$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

$$t=x-\frac{1}{x}dt=1+\frac{1}{x^2}dt$$$$\int {tan}^{-1}dt$$$$t\times {tan}^{-1}t-\int \frac{1}{1+t^2}\times tdt$$$$t\times {tan}^{-1}t-\frac{1}{2}ln(1+t^2)+c$$$$(x-\frac{1}{x}){tan}^{-1}(x-\frac{1}{x})-\frac{1}{2}ln\mid (1+x^2+\frac{1}{x^2}-2)\mid +c$$$$(x-\frac{1}{x}){tan}^{-1}(x-\frac{1}{x})-\frac{1}{2}ln\mid x^2+\frac{1}{x^2}-1\mid +c$$

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