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Evaluate $\int\frac{4x+1}{x^{2}-36}dx$
Option: 1
$\frac{23}{12}ln\left | x-6 \right |-\frac{25}{12}ln\left | x+6 \right |+C$

Option: 2
$\frac{23}{12}ln\left | x-6 \right |+\frac{25}{12}ln\left | x+6 \right |+C$

Option: 3
$\frac{23}{12}ln\left | x+6 \right |-\frac{25}{12}ln\left | x-6 \right |+C$

Option: 4
$\frac{23}{12}ln\left | x+6 \right |+\frac{25}{12}ln\left | x-6 \right |+C$

Answers (1)

best_answer

Given integral,

$\int\frac{4x+1}{x^{2}-36}dx$

$\int\frac{4x+1}{x^{2}-36}dx=\int\frac{4x+1}{\left ( x+6 \right )\left ( x-6 \right )}dx$

$\int\frac{4x+1}{x^{2}-36}dx=\int\left ( \frac{A}{x+6}+\frac{B}{x-6} \right )dx$

Using partial fractions,

$A\left ( x-6 \right )+B\left ( x+6 \right )=4x+1$

Put $x=-6,$

$-12A=-24+1\Rightarrow A=\frac{23}{12}$

Put $x=6,$

$12B=24+1\Rightarrow B=\frac{25}{12}$

Now,

$\int\frac{4x+1}{x^{2}-36}dx=\int\left ( \frac{\frac{23}{12}}{x+3}+\frac{\frac{25}{12}}{x-3} \right )dx$

$\int\frac{4x+1}{x^{2}-36}dx=\int\left ( \frac{23}{12}\frac{1}{x+6}+\frac{25}{12}\frac{1}{x-6} \right )dx$

$\int\frac{4x+1}{x^{2}-36}dx=\frac{23}{12}ln\left | x+6 \right |+\frac{25}{12}ln\left | x-6 \right |+C$

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