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Question Number 98677 by PRITHWISH SEN 2 last updated on 15/Jun/20

$$\mathrm{prove}$$ $${\int }_0^{\mathrm{a}}\frac{\mathbf{ln}(1+\mathbf{ax})}{1+{\mathbf{x}}^2}\mathbf{dx}=\frac{1}{2}\mathbf{ln}(1+{\mathbf{a}}^2){\tan }^{-1}\mathbf{a},\mathbf{a}>0$$

Answered by maths mind last updated on 15/Jun/20

$$f\left(t\right)={\int }_0^a\frac{ln(1+tx)}{(1+x^2)}$$ $$f^{\prime }\left(t\right)={\int }_0^a\frac{xdx}{(1+tx)(1+x^2)}$$ $$={\int }_0^a\frac{(x+t)(1+tx)-t(1+x^2)}{(1+t^2)(1+tx)(1+x^2)}$$ $$={\int }_0^a\frac{x+t}{(1+t^2)(1+x^2)}dx-{\int }_0^a\frac{tdx}{(1+t^2)(1+tx)}$$ $$=\frac{ln(1+a^2)}{2(1+t^2)}+\frac{tarctan\left(a\right)}{t^2+1}-\frac{ln(1+ta)}{1+t^2}$$ $$f\left(0\right)=0$$ $$f\left(a\right)={\int }_0^a\frac{ln(1+ta)}{1+t^2}dt$$ $$f\left(a\right)={\int }_0^{a}f\left(t\right)dt\Rightarrow f\left(a\right)=\int (\frac{ln(1+a^2)}{2(1+t^2)}+\frac{t}{t^2+1}arctan\left(a\right)-\frac{ln(1+ta)}{1+t^2})dt$$ $$\Rightarrow 2f\left(a\right)={\int }_0^a(\frac{ln(1+a^2)}{2(1+t^2)}+\frac{t}{t^2+1}arctan\left(a\right))dt$$ $$=\frac{ln(1+a^2)}{2}arctan\left(a\right)+\frac{arctan\left(a\right)}{2}ln(1+a^2)$$ $$\iff f\left(a\right)=\frac{ln(1+a^2)}{2}{\tan }^{-1}\left(a\right)$$

Commented byPRITHWISH SEN 2 last updated on 16/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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