Find ∫ (2x^2+1)e^x^2 dx

Question

Does this problem have many solutions?

Answer 1

THe anti-derivative is found as follows:

1st IBP:

U = 2x^2 + 1 dV = (exp(x))^2 = e^(2x)

dU = 4x V = (1/2) exp(2x)

(2x^2+1)(1/2) exp(2x) - 2 integral [ x * exp(2x)]

2nd IBP:

U = x dV = exp(2x)

dU = dx V = (1/2)exp(2x)

(2x^2+1)(1/2) exp(2x) - [ x*exp(2x) - integral (exp(2x)) ] <-- 2 cancels 1/2

(2x^2+1)(1/2) exp(2x) - x*exp(2x) + integral (exp(2x)) <-- distributes negative

(2x^2+1)(1/2) exp(2x) - x*exp(2x) + (1/2)exp(2x) <--- integrates

(x^2 -x + 1 ) exp(2x) <--- factors out exp(2x) and combines like terms

So yes, the GENERAL anti-derivative is technically

(x^2 -x + 1 ) exp(2x) + C , where C is ANY fixed # constant