If ∫(x+√1+x2)ndx=1a(n+1){x+√1+x2}n+1+1b(n−1){x+√1+x2}n−1+C;(n>1) then a+b=

If ∫(x+√1+x2)ndx=1a(n+1){x+√1+x2}n+1+1b(n−1){x+√1+x2}n−1+C

1.	If ∫(x+√1+x2)ndx=1a(n+1){x+√1+x2}n+1+1b(n−1){x+√1+x2}n−1+C;(n>1) then a+b=
Answer» If $\int (x + \sqrt{1 + x^{2}})^{n} d x = \frac{1}{a (n + 1)} {x + \sqrt{1 + x^{2}}}^{n + 1} + \frac{1}{b (n - 1)} {x + \sqrt{1 + x^{2}}}^{n - 1} + C; (n > 1)$ then $a + b =$