Question

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

Answer 1

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.