$$ f(x)=\int e^{x^4}{(1+x^2+2x^4)}\,d(e^{x^2}) $$ Find $f(1)+f’(0)$, if $f(0) = 0$.
I tried assuming $e^{x^2}$ as a dummy variable $u$, converting everything else in terms of natural log: $$ f(x) = \int (\ln(u))^2 \bigl[1 + \ln(u) + 2(\ln(u))^2\bigr]\,du $$
But can't go forward from this
Edit
Later understood that $e^{x^4} \neq (e^{x^2})^2$
On
Hint: Suppose $f(x)=p(x)\cdot e^{x^2+x^4}$ (so $p(0)=0$) for some polynomial function $p(x)$, so that upon differentiation we have
$$f'(x) = \left[p'(x) + p(x) (2x+4x^3) \right] e^{x^2+x^4}= 2x (1+x^2+2x^4) e^{x^2+x^4}$$
Finding $p(x)$, if it exists, is now a matter of solving the linear ODE
$$p'(x) + 2x(1+2x^2) p(x) = 2x(1+x^2+2x^4)$$
On
Given a function $g(x)$ that is $C^\infty$ (smooth) over $\mathbb R$, then a Riemann–Stieltjes integral is equivalent to the Riemann integral of $$\int_a^b f(x) \text{ d}g(x) = \int_a^b f(x)g'(x)\text{ d}x$$
Your integral $$\int e^{x^4}{(1+x^2+2x^4)}\,d(e^{x^2})$$ has $g(x) = e^{x^2}$, which is smooth, so we can transform your integral into \begin{align} \int e^{x^4}{(1+x^2+2x^4)}\cdot 2xe^{x^2}\text{ d}x & = \int e^{x^2(x^2+1)} (2x^4 + x^2 + 1) \cdot 2x \text{ d}x\\ &= \int e^{u(u+1)}(2u^2 + u + 1) \text{ d}u, \qquad u=x^2 \end{align}
Expand out the integrand and integrate the third term by parts, which cancels out the second, leaving the first one. Hence, the antiderivative is $x^2 e^{x^4 + x^2} + C$. If $f(0) = 0$, then $C=0$.
We now have $f(x) = x^2 e^{x^4 + x^2}$, and evaluating $f(1)+f’(0)$ should be trivial.
Rewrite the integral as $$ I=\int e^{x^4}\left(1+x^2+2 x^4\right) d\left(e^{x^2}\right) = \int e^{x^4}\left(1+x^2+2 x^4\right) e^{x^2} d\left(x^2\right) $$ Letting $y=x^2$ yields $$ \begin{aligned} I & =\int e^{y^2+y}\left(1+y+2 y^2\right) d y \\ & =\int e^{y^2+y}[y(2 y+1)+1] d y \\ & =\int y d\left(e^{y^2+y}\right)+\int e^{y^2+y} d y \\ & =y e^{y^2+y}-\int e^{y^2+y} d y+\int e^{y^2+y} d y \\ & =y e^{y^2+y}+C\\ & =x^2 e^{x^4+x^2}+C \end{aligned} $$