Question

$\int \left( x + \frac { 1 } { x } \right) ^ { n + 5 } \left( \frac { x ^ { 2 } - 1 } { x ^ { 2 } } \right) d x$ is equal to

• A.
• B.
• C. $\left( \frac { x } { x ^ { 2 } + 1 } \right) ^ { n + 6 } ( n + 6 ) + c$
• D. none of these

Answer 1

Answer: (A)

Answer 2

Here I =.

If$x + \frac { 1 } { x } = p$ then, ,

⇒ I =$\int \left( x + \frac { 1 } { x } \right) ^ { n + 5 } \left( \frac { x ^ { 2 } - 1 } { x ^ { 2 } } \right) d x = \int p ^ { n + 5 } d p$

$= \frac { p ^ { n + 6 } } { n + 6 } + c$