Question

Evaluate the integral.?

Answer 1

`int_0^1 (1+x^2)^3*dx=96/35`

Explanation:

`int_0^1 (1+x^2)^3*dx`

`=int_0^1 (x^6+3x^4+3x^2+1)*dx`

`=[x^7/7+(3x^5)/5+x^3+x]_0^1`

=`1/7+3/5+1+1`

=`96/35`