In these type of problems , the aim is to part - integrate the integral in such a way that the second function is the one whose " value " , i . e , in this case , the power of " t m " is seen to have increased . In the solution below , the limits are from " 0 " to " 1 " .
I ( m , n ) = ∫ t m ( 1 + t ) n dt
= ( 1 + t ) n ∫ t m dt - ∫ d ( 1 + t ) ndt [ ∫ t m dt ] dt
= [ t m + 1 ( 1 + t ) nm + 1 ] - nm + 1 ∫ t m + 1 ( 1 + t ) n - 1 dt
= 2 nm + 1 - nm + 1 I ( m + 1 , n - 1 )........................After putting the limits .
Hence , the answer is " ( a ) " .