d x x^2-9 equal to : - Vidyadip

MCQ

$\int \frac{d x}{x^2-9}$ equal to :

✓
$\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+C$
B
$\frac{1}{6} \log \left(\frac{x+3}{x-3}\right)+C$
C
$\frac{1}{3} \log \left(\frac{x-3}{x+3}\right)+C$
D
$\frac{1}{3} \log \left(\frac{x+3}{x-3}\right)+ C$

✓

Answer

Correct option: A.

$\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+C$

(A)
$
\begin{aligned}
I= & \int \frac{d x}{x^2-9}=\int \frac{d x}{(x)^2-(3)^2} \\
& \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+C \\
\therefore \quad I & =\frac{1}{2 \times 3} \log \left(\frac{x-3}{x+3}\right)+C \\
I & =\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+C
\end{aligned}
$
Hence option (A) is correct.

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