d x x^2-9 बराबर है-

MCQ

$\int \frac{d x}{x^2-9}$ बराबर है$-$

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

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

$\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+ C$
$I =\int \frac{d x}{x^2-9}=\int \frac{d x}{(x)^2-(3)^2}$
$\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)+ C$
$\therefore \quad I =\frac{1}{2 \times 3} \log \left(\frac{x-3}{x+3}\right)+ C$
$I =\frac{1}{6} \log \left(\frac{x-3}{x+3}\right)+ C$
अतः सही विकल्प $(A)$ है।

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