Q: $\int ( x + \frac{1}{x})^{n+5} (\frac{x^2 -1}{x^2})dx $ is equal to
(A) $\frac{(x+ \frac{1}{x})^{n+6}}{n+6} + c $
(B) $ (\frac{x^2 + 1}{x^2})^{n+6} (n+6) + c $
(C) $ (\frac{x}{x^2 + 1})^{n+6} (n+6) + c $
(D) none of these
Sol: Let $(x+ \frac{1}{x}) = z $ then $(1 – \frac{1}{x^2})dx = dz $
$I = \int z^{n+5} dz = \frac{z^{n+6}}{n+6} + c $
$ = \frac{(x + \frac{1}{x})^{n+6}}{n+6} + c $
Hence (A) is the correct answer.