Relation on $\int_1^x\exp{t^2}dt$

Question

Could you give me some leads to show the following relation : $$\forall x>0, \int_1^x\exp{t^2}dt = \frac{1}{2x}\exp{x^2} + \frac{1}{4x^3}\exp{x^2} - \frac{3}{4}\mathbb{e}+ \frac{3}{4} \int_1^x \frac{1}{t^4}\exp{t^2}dt $$

Thanks in advance.

Answer 1

Hint 1: multiply the integrand by $1 = \frac{t}{t}$, then use parts.

Hint 2: in parts, let $dv = t e^{t^2}dt$ always; an antiderivative is $\frac{1}{2} e^{t^2}$.

Hint 3: do this twice.

Answer 2

A straighforward way is to compute the derivative (in $x$) of the two members of the egality, noticing for $x=1$ that both are $0$. The result is all these derivatives are null function.