\int \frac{dx}{(x^2-9)} equals ______

(a) \frac{1}{6} log \frac{x+3}{x-3} + C

(b) \frac{1}{6} log \frac{x-3}{x+3} + C

(c) \frac{1}{5} log \frac{x+3}{x-3} + C

(d) \frac{1}{3} log \frac{x+3}{x-3} + C

This question was addressed to me in an interview for job.

My question is from Integration by Partial Fractions in section Integrals of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

The correct answer is (b) \frac{1}{6} log \frac{x-3}{x+3} + C

The explanation is: \int \frac{dx}{(x^2-9)}=\frac{A}{(x-3)} + \frac{B}{(x+3)}

By simplifying, it we get \frac{A(x+3)+B(x-3)}{(x^2-9)} = \frac{(A+B)x+3A-3B}{(x^2-9)}

By solving the equations, we get, A+B=0 and 3A-3B=1

By solving these 2 equations, we get values of A=1/6 and B=-1/6.

Now by putting values in the equation and integrating it we get value,

\frac{1}{6} log (\frac{x-3}{x+3}) + C.