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Question Number 212023 by Spillover last updated on 27/Sep/24

Answered by Frix last updated on 27/Sep/24

$$\int \frac{(x+1)\tan x}{{(1+\tan x)}^2}dx=$$$$=\frac{1}{2}\int (x+1-\frac{1}{\underset{[t=\tan x]}{\underbrace{1+\sin 2x}}}-\frac{x}{\underset{\left[\mathrm{by}\mathrm{parts}\right]}{\underbrace{1+\sin 2x}}})dx=$$$$...$$$$=\frac{x(x+1)}{4}+\frac{(x+1)}{2(1+\tan x)}+\frac{1}{8}\ln \frac{1+{\tan }^{2}x}{{(1+\tan x)}^2}+C$$$$\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\frac{(x+1)\tan x}{{(1+\tan x)}^2}dx=\frac{{\pi }^2}{64}+\frac{\pi }{8}-\frac{1}{4}-\frac{\ln 2}{8}$$$$$$$$[\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{(x+1)\tan x}{{(1+\tan x)}^2}dx=\frac{{\pi }^2}{16}+\frac{\pi }{8}-\frac{1}{2}]$$

Commented by Spillover last updated on 27/Sep/24

$$correct$$

Answered by Spillover last updated on 27/Sep/24

Answered by Spillover last updated on 27/Sep/24

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