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### Answers (1)

Answer:

$I=\frac{1}{\sqrt{17}} \log \left|\frac{8 x+1+\sqrt{17}}{-8 x-1+\sqrt{17}}\right|+c$

Given:

$\int \frac{1}{1-x-4 x^{2}} d x$

Hint:

To solve this statement we have to do whole square method.

Solution:

$\int \frac{1}{1-4\left(\frac{x}{4}+x^{2}\right)} d x$

$I=\int \frac{1}{1-4\left(x+\frac{1}{8}\right)^{2}+\frac{1}{16}} d x$

$I=\int \frac{1}{\frac{17}{6}-4\left(x+\frac{1}{8}\right)^{2}} d x$

$I=\frac{1}{4} \int \frac{1}{\left(\frac{\sqrt{17}}{8}\right)^{2}-\left(x+\frac{1}{8}\right)^{2}} d x$                            $\left[\because \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right]$

$I=\frac{1}{4} \frac{4}{\sqrt{17}} \log \left|\frac{x+\frac{1}{8}+\frac{\sqrt{17}}{8}}{\frac{\sqrt{17}}{8}-x-\frac{1}{8}}\right|+c$

$I=\frac{1}{\sqrt{17}} \log \left|\frac{8 x+1+\sqrt{17}}{-8 x-1+\sqrt{17}}\right|+c$

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