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Question Number 27539 by Mr eaay last updated on 08/Jan/18

Commented by Tinkutara last updated on 08/Jan/18

For second part see question 27445 and 27400.

Answered by Joel578 last updated on 09/Jan/18

$$I=\int \frac{x+2}{\sqrt{x^2+9}}dx=\int \frac{x}{\sqrt{x^2+9}}dx+2\int \frac{1}{\sqrt{x^2+9}}dx$$$$$$$$I_1=\int \frac{x}{\sqrt{x^2+9}}dx$$$$\mathrm{Let}u=x^2+9\to du=2xdx$$$$I_1=\int \frac{x}{\sqrt{u}}.\frac{du}{2x}=\frac{1}{2}\int \frac{1}{\sqrt{u}}du$$$$=\sqrt{u}+C=\sqrt{x^2+9}+C$$$$$$$$I_2=\int \frac{1}{\sqrt{x^2+9}}dx$$$$\mathrm{Let}x=3\tan \theta \to dx={3\sec }^2\theta d\theta$$$$I_2=\int \frac{1}{\sqrt{{9\tan }^2\theta +9}}.{3\sec }^2\theta d\theta$$$$=\int \frac{1}{3\sec \theta }.{3\sec }^2\theta d\theta =\int \sec \theta d\theta$$$$=\ln \mid \sec \theta +\tan \theta \mid +C$$$$=\ln \mid \frac{1}{3}(\sqrt{x^2+9}+x)\mid +C$$$$$$$$I=I_1+2I_2$$$$I=\sqrt{x^2+9}+2\ln \mid \frac{1}{3}(\sqrt{x^2+9}+x)\mid +C$$

Answered by Joel578 last updated on 09/Jan/18

$$\int \frac{x-8}{x^2+4x+16}dx=\int \frac{x+2-10}{{(x+2)}^2+12}dx=\int \frac{x+2}{{(x+2)}^2+12}dx-10\int \frac{1}{{(x+2)}^2+12}dx$$$$$$$$I_1=\int \frac{x+2}{{(x+2)}^2+12}dx$$$$\mathrm{Let}u=x+2\to du=dx$$$$I_1=\int \frac{u}{u^2+12}du=\frac{1}{2}\ln (u^2+12)+C=\frac{1}{2}\ln ({(x+2)}^2+12)+C$$$$$$$$I_2=\int \frac{1}{{(x+2)}^2+12}dx$$$$\mathrm{Let}u=x+2\to du=dx$$$$I_2=\int \frac{1}{u^2+{\left(\sqrt{12}\right)}^2}du=\frac{1}{\sqrt{12}}{\tan }^{-1}\left(\frac{u}{\sqrt{12}}\right)+C$$$$=\frac{\sqrt{3}}{6}{\tan }^{-1}\left(\frac{(x+2)\sqrt{3}}{6}\right)+C$$$$$$$$I=I_1-10I_2$$$$I=\frac{1}{2}\ln ({(x+2)}^2+12)-\frac{5\sqrt{3}}{3}{\tan }^{-1}\left(\frac{(x+2)\sqrt{3}}{6}\right)+C$$

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