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Evaluate the integral of $\int\frac{x^{2}-1}{x^{2}-9}dx$

Option: 1
$x+\frac{4}{3}ln\frac{\left | x-3 \right |}{\left | x+3 \right |}+C$

Option: 2
$x-\frac{4}{3}ln\frac{\left | x-3 \right |}{\left | x+3 \right |}+C$

Option: 3
$x+\frac{4}{3}ln\frac{\left | x+3 \right |}{\left | x-3 \right |}+C$

Option: 4
$x-\frac{4}{3}ln\frac{\left | x+3 \right |}{\left | x-3 \right |}+C$

Answers (1)

best_answer

Given integral,

$\int\frac{x^{2}-1}{x^{2}-9}dx$

$\int\frac{x^{2}-1}{x^{2}-9}dx=\int\left ( 1+\frac{8}{\left ( x+3 \right )\left ( x-3 \right )} \right )dx$

$\int\frac{x^{2}-1}{x^{2}-9}dx=\int\left ( 1+\frac{A}{x+3}+\frac{B}{x-3} \right)dx$

Using partial fractions,

$A\left ( x-3 \right )+B\left ( x+3 \right )=8$

Put $x=3,$

$B=\frac{4}{3}$

Put $x=-3$

$A=-\frac{4}{3}$

Now,

$\int\frac{x^{2}-1}{x^{2}-9}dx=\int\left ( 1+\frac{-\frac{4}{3}}{x+3}+\frac{\frac{4}{3}}{x-3} \right)dx$

$\int\frac{x^{2}-1}{x^{2}-9}dx=x-\frac{4}{3}ln\left | x+3 \right |+\frac{4}{3}ln\left | x-3 \right |+C$

$\int\frac{x^{2}-1}{x^{2}-9}dx=x+\frac{4}{3}\left ( ln\left | x-3 \right |-ln\left |x+3 \right | \right )+C$

$\int\frac{x^{2}-1}{x^{2}-9}dx=x+\frac{4}{3}ln\frac{\left | x-3 \right |}{\left | x+3 \right |}+C$

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