1.

\(\int \frac{dx}{(x^2-9)}\) 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) \(\frac{1}{3} log \frac{x+3}{x-3}\) + CThis 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»

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.


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