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Question Number 191009 by pascal889 last updated on 16/Apr/23

Answered by Frix last updated on 16/Apr/23

$$\int \frac{3x+2}{x^2+6x+4}dt=\frac{3x+2}{x^2+6x+4}\int dt=\frac{3x+2}{x^2+6x+4}t+C$$

Commented by ARUNG_Brandon_MBU last updated on 16/Apr/23

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Answered by Frix last updated on 16/Apr/23

$$\int \frac{3x+2}{x^2+6x+4}dx=$$$$=\int (\frac{15+7\sqrt{5}}{10(x+3+\sqrt{5})}+\frac{15-7\sqrt{5}}{10(x+3-\sqrt{5})})dx=$$$$=\frac{(15+7\sqrt{5})\ln \mid x+3+\sqrt{5}\mid +(15-7\sqrt{5})\ln \mid x+3-\sqrt{5}\mid }{10}+C$$

Answered by ARUNG_Brandon_MBU last updated on 16/Apr/23

$$I=\int \frac{3x+2}{x^2+6x+4}$$$$=\frac{3}{2}\int \frac{2x+6}{x^2+6x+4}dx-\int \frac{7dx}{x^2+6x+4}$$$$=\frac{3}{2}\ln \mid x^2+6x+4\mid -\int \frac{7dx}{{(x+3)}^2-5}$$$$=\frac{3}{2}\ln \mid x^2+6x+4\mid +\frac{7}{\sqrt{5}}\mathrm{argcoth}\left(\frac{x+3}{\sqrt{5}}\right)+C$$$$=\frac{3}{2}\ln \mid x^2+6x+4\mid +\frac{7\sqrt{5}}{10}\ln \mid \frac{x+3+\sqrt{5}}{x+3-\sqrt{5}}\mid +C$$

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