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### Answers (1)

Answer:

$\frac{1}{17} \log \left|\frac{x-\frac{2}{8}}{x+5}\right|+c$

Given:

$\int \frac{1}{3 x^{2}+13 x-10} d x$

Hint:

In this equation we have to make whole square statement from quadratic equation.

Solution:

$I=\frac{1}{3} \int \frac{1}{x^{2}+\frac{18 x}{8}-\frac{10}{8}} d x$

$I=\frac{1}{3} \int \frac{1}{\left(x+\frac{18}{6}\right)^{2}-\frac{169}{36}-\frac{10}{3}} d x$

$I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{18}{6}\right)^{2}-\frac{169+120}{36}}$

$I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{18}{6}\right)^{2}-\frac{289}{36}}$

$I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}$

$I=\frac{1}{3} \frac{1}{2 \times \frac{17}{6}} \log \left|\frac{x+\frac{18}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+c$

$I=\frac{1}{17}+\log \left|\frac{x-\frac{4}{6}}{x+\frac{30}{6}}\right|+c$

$I=\frac{1}{17} \log \left|\frac{x-\frac{2}{8}}{x+5}\right|+c$

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