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Question Number 33643 by mondodotto@gmail.com last updated on 21/Apr/18

$$\int \frac{2\mathit{x}+3}{{\mathit{x}}^2+4}\mathit{d}\mathit{x}$$

Answered by Joel578 last updated on 21/Apr/18

$$I=\int \frac{2x}{x^2+4}dx+3\int \frac{1}{x^2+4}dx$$$$u=x^2+4\to du=2xdx$$$$$$$$I=\int \frac{2x}{u}\left(\frac{du}{2x}\right)+\frac{3}{2}{\tan }^{-1}\left(\frac{x}{2}\right)$$$$=\ln (x^2+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C$$

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