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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

Posted by

Rishabh

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