Q.If $\int f'(x)e^{x^2}dx=(x-1)e^{x^2}+c$, then $f(x)$ is

Given $\int f(x)\,dx=g(x)+c$, differentiating shows $g'(x)=f(x)$, so $\int f(x)g'(x)\,dx=\int (f(x))^2\,dx$.

Given $\int f(x)\,dx=g(x)+c$. Differentiating both sides with respect to $x$: $f(x)=g'(x)$.

Substitute $g'(x)=f(x)$ into $\int f(x)\,g'(x)\,dx$:

$\int f(x)\,g'(x)\,dx=\int f(x)\cdot f(x)\,dx=\int (f(x))^2\,dx$

Differentiating both sides of the given equation recovers $f'(x)=2x^2-2x+1$, and integrating that gives $f(x)=\dfrac{2x^3}{3}-x^2+x+c$.

Since $\int f'(x)e^{x^2}\,dx = (x-1)e^{x^2}+c$, differentiating both sides with respect to $x$ must give back the integrand:

$f'(x)e^{x^2} = \dfrac{d}{dx}\big[(x-1)e^{x^2}\big] = e^{x^2} + (x-1)(2x)e^{x^2}$ (product rule, using $\frac{d}{dx}e^{x^2}=2xe^{x^2}$)

$= e^{x^2}\big[1+2x(x-1)\big] = e^{x^2}(2x^2-2x+1)$.

Dividing both sides by $e^{x^2}$: $f'(x) = 2x^2-2x+1$.

Integrating: $f(x) = \displaystyle\int(2x^2-2x+1)\,dx = \dfrac{2x^3}{3}-x^2+x+c$.