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Question Number 187703 by Tawa11 last updated on 20/Feb/23

$$\int \frac{1}{{5\mathrm{x}}^2-2\mathrm{x}-4}\mathrm{dx}$$

Answered by MikeH last updated on 20/Feb/23

$$=\frac{1}{5}\int \frac{1}{x^2-\frac{2}{5}x-\frac{4}{5}}dx$$$$=\frac{1}{5}\int \frac{1}{{(x-\frac{1}{5})}^2-\frac{1}{25}-\frac{4}{5}}dx$$$$=\frac{1}{5}\int \frac{1}{{(x-\frac{1}{5})}^2-\frac{21}{25}}dx=\frac{1}{5}\int \frac{1}{{(x-\frac{1}{5})}^2-{\left(\frac{\sqrt{21}}{5}\right)}^2}dx$$$$=\frac{1}{5}\int \frac{1}{u^2-a^2}du$$$$\mathrm{with}u=(x-\frac{1}{5})$$$$I=\frac{1}{5}\int \frac{1}{u^2-a^2}du=\frac{1}{10a}\ln \left(\frac{u-a}{u+a}\right)+c$$$$\Rightarrow I=\frac{1}{2\sqrt{21}}\ln \left(\frac{x-\frac{1+\sqrt{21}}{5}}{x+\frac{1-\sqrt{21}}{5}}\right)+c$$$$I=\frac{1}{2\sqrt{21}}\ln \left(\frac{5x-(1+\sqrt{21})}{5x+(1-\sqrt{21})}\right)+c$$$$\mathrm{please}\mathrm{correct}\mathrm{me}\mathrm{if}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{wrong}.$$$$$$

Commented by Ar Brandon last updated on 20/Feb/23

$$\Rightarrow I=\frac{1}{2\sqrt{21}}\ln \mid \frac{x-\frac{1+\sqrt{21}}{5}}{x-\frac{1-\sqrt{21}}{5}}\mid +c$$

Answered by Nimnim111118 last updated on 21/Feb/23

$$I=\int \frac{5dx}{25x^2-10x-20}=\int \frac{5dx}{{(5x-1)}^2-{\left(\sqrt{21}\right)}^2}$$$$=\frac{1}{2\sqrt{21}}ln\left(\frac{5x-1-\sqrt{21}}{5x-1+\sqrt{21}}\right)+C$$

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