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Question Number 210354 by klipto last updated on 08/Aug/24

$${\int }_0^{\mathit{\alpha }}\frac{\mathbf{x}}{(1+{\mathbf{x}}^2)(1+\mathit{\alpha }\mathbf{x})}\mathbf{dx}$$

Answered by klipto last updated on 08/Aug/24

Answered by Frix last updated on 08/Aug/24

$$\int \frac{x}{(x^2+1)(\alpha x+1)}dx=$$$$=\frac{1}{{\alpha }^2+1}\int \frac{x}{x^2+1}dx+\frac{\alpha }{{\alpha }^2+1}\int \frac{dx}{x^2+1}-\frac{\alpha }{{\alpha }^2+1}\int \frac{dx}{\alpha x+1}=$$$$=\frac{\ln (x^2+1)}{2({\alpha }^2+1)}+\frac{\alpha {\tan }^{-1}x}{{\alpha }^2+1}-\frac{\ln \mid \alpha x+1\mid }{{\alpha }^2+1}+C$$$$\underset{0}{\stackrel{\alpha }{\int }}\frac{x}{(x^2+1)(\alpha x+1)}dx=$$$$=\frac{\alpha {\tan }^{-1}\alpha }{{\alpha }^2+1}-\frac{\ln ({\alpha }^2+1)}{2({\alpha }^2+1)}$$

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