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Mathematics Question on Integration by Parts

If I(m,n) =∫01tm(1+t)ndt,\int_0^1 t^m(1+t)n dt , then the expression for I(m, n) in terms of I(m + 1, n - 1) is

A

2nm+1−nm+1(m+1,n−1)\frac{2^n}{m+1}-\frac{n}{m+1}(m+1,n-1)

B

nm+1I(m+1,n−1)\frac{n}{m+1}I(m+1,n-1)

C

2nm+1+nm+1I(m+1,n−1)\frac{2^n}{m+1}+\frac{n}{m+1}I(m+1,n-1)

D

mm+1I(m+1,n−1)\frac{m}{m+1}I(m+1,n-1)

Answer

2nm+1−nm+1(m+1,n−1)\frac{2^n}{m+1}-\frac{n}{m+1}(m+1,n-1)

Explanation

Solution

The correct option is:(A) 2nm+1−nm+1(m+1,n−1)\frac{2^n}{m+1}-\frac{n}{m+1}(m+1,n-1).

Here, I(m,n) =∫01tm(1+t)ndt,\int_0^1 t^m(1+t)n dt , reduce into I(m + 1, n - 1)
[we apply integration by parts taking (1 + t)n^n as first
and tm^m as second function]
∴     i(m,n)=[(1+t)n.tm+1m+1]01 −∫01n(1+t)n−1.tm+1m+1dt\therefore \, \, \, \, \, i(m,n)=\bigg[(1+t)^n.\frac{t^{m+1}}{m+1}\bigg]_0^1 \, -\int_0^1n(1+t)^{n-1}.\frac{t^{m+1}}{m+1}dt
        =2nm+1−nm+1∫01(1+t)(n−1),tm+1dt\, \, \, \, \, \, \, \, =\frac{2^n}{m+1}-\frac{n}{m+1} \int_0^1 (1+t)^{(n-1)},t^{m+1}dt
∴  I(m,n)=2nm+1−nm+1.I(m+1,n−1)\therefore \, \, I(m,n) =\frac{2^n}{m+1}-\frac{n}{m+1}.I(m+1,n-1)