Question

How do you integrate `int 3xe^(x^2+1) dx`?

Answer 1

In this case, the integrand is the derivative of a function composition (up to a multiplicative constant):
`g(y)=e^y`
`f(x)=x^2+1`
`rArr g(f(x))=e^(x^2+1)`
`rArr d/(dx) [g(f(x))]=d/(dx)[e^(x^2+1)]=2x e^(x^2+1)`

The only difference is the multiplicative constant (it's `2` instead of `3`), so we can rewrite the integral to look like the derivative of `e^(x^2+1)` by working with multiplicative constants:
`int 3x e^(x^2+1) dx=3/2 int 2x e^(x^2+1) dx=3/2 e^(x^2+1) + C`

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Another (very similar) way of solving this was to consider the substitution `u(x)=x^2+1`. We could relate the two differentials `du=2x dx` and rewrite the integral in terms of `u`:
`int 3x e^(x^2+1) dx=int 3/2 e^(x^2+1) (2xdx)=int 3/2 e^u du=3/2e^u+C`
Now we could return back to `x`:
`3/2e^u+C=3/2 e^(x^2+1)+C`