Evaluate `int (1+x)^2 dx`

Explanation

Method I :
`1 xx 2 xx .^(n)C_(1) + 2 xx 3 xx .^(n)C_(2) + "...." + n xx (n+1) xx .^(n)C_(n)`
`= underset(r=1)overset(n)sumr(r+1)..^(n)C_(r)`
`= underset(r=1)overset(n)sum(r+1)[n..^(n-1)C_(r-1)]`
`= n underset(r=1)overset(n)sum((r-1)+2).^(n-1)C_(r-1)`
`= nxxunderset(r=1)overset(n)sum[(r-1)..^(n-1)C_(r-1)+2..^(n-1)C_(r-1)]`
`= nxx n(n-1)underset(r=2)overset(n)sum.^(n-2)C_(r-2)+(2n)underset(r=1)overset(n)sum.^(n-1)C_(r-1)`
`= n xx (n-1) xx 2^(n-2) + 2n xx 2^(n-1)`
` = n(n+3)xx2^(n-2)`
Method II :
We have `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x + .^(n)C_(2)x^(2) + "....." + .^(n)C_(n)x^(n)`
Differentiating w.r.t.x we get
`n(1+x)^(n-1) = .^(n)C_(1) + 2 xx .^(n)C_(2)x + 3 xx .^(n)C_(3)x^(2)"....." + nxx .^(n)C_(n) xx x^(n-1)`
Multipying with `x^(2)`, we have
`n(1+x)^(n-1) x^(2) = .^(n)C_(1) x^(2) + 2 xx .^(n)C_(2)x^(3) + 3 xx .^(n)C_(3)x^(4)"...."+n xx .^(n)C_(n)x^(n+1)`
Differentiating w.r.t. , we get
`m(n-1)(1+x)^(n-2)x^(2) + 2x xx n(1+x)^(n-1)`
`= 2.^(n)C_(1)x + 2 xx 3 xx .^(n)C_(2)x^(2) + 3 xx 4 xx .^(n)C_(3)x^(3) + "...." + n(n+1) xx .^(n)C_(n)x^(n)`
Now putting ` x = 1`, we get
`1 xx 2 xx .^(n)C_(1) + 2 xx 3 xx .^(n)C_(2) + "...." + n xx (n+1) xx .^(n)C_(n)`
`= n (n+3) xx 2^(n-2)`