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Question Number 20908 by j.masanja06@gmail.com last updated on 07/Sep/17

$$integratewithrespecttox$$$$\int \left(\frac{2x+1}{x^2+4x+8}\right)dx$$

Answered by Joel577 last updated on 07/Sep/17

$$I=\int \frac{2x+1}{{(x+2)}^2+4}dx$$$$$$$$\mathrm{Let}u=x+2\to du=dx$$$$I=\int \frac{2u-3}{u^2+4}du$$$$$$$$\mathrm{Let}u=2\tan \theta \to du={2\sec }^2\theta d\theta$$$$\tan \theta =\frac{u}{2}\to \theta ={\tan }^{-1}\left(\frac{u}{2}\right)$$$$I=\int \frac{4\tan \theta -3}{4({\tan }^2\theta +1)}{2\sec }^2\theta d\theta$$$$=\int \frac{4\tan \theta -3}{2}d\theta =\int 2\tan \theta -\frac{3}{2}d\theta$$$$=-2\ln \mid \cos \theta \mid -\frac{3}{2}\theta +C$$$$=-2\ln \mid \frac{2}{\sqrt{u^2+4}}\mid -\frac{3}{2}{\tan }^{-1}\left(\frac{u}{2}\right)+C$$$$=-2\ln \mid \frac{2}{\sqrt{{(x+2)}^2+4}}\mid -\frac{3}{2}{\tan }^{-1}\left(\frac{x+2}{2}\right)+C$$

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