How do you integrate $intx\(3^(x^2+1))dx$ ?

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Answer 1 · 2018-03-22T08:03:32

The answer is $=1/2*3^(x^2+1)/ln3 +C$

Let $u=x^2+1$ , $=>$ , $du=2xdx$

Therefore,

$I=intx(3^(x^2+1))dx=1/2int3^udu$

Let $v=3^u$

$lnv=ln(3^u) = u ln3$

$v=e ^(u ln3)$

Therefore,

$I=1/2inte ^(u ln3)du$

$=1/2*e ^(u ln3)/(ln3)$

$=1/2*3^(x^2+1)/ln3 +C$

How do you integrate intx\(3^(x^2+1))dxintx\(3^(x^2+1))dx?