Integration by substitution

Idea. Spot an inner function $u(t)$ whose derivative appears (up to a constant) elsewhere in the integrand. Do not forget to change the bounds.

First integral. With $u=t^2$, $du=2t\,dt$, and the bounds become $0$ and $1$:
$$\int_0^1\frac t{1+t^4}\,dt=\frac12\int_0^1\frac{du}{1+u^2}=\frac12\bigl[\arctan u\bigr]_0^1=\frac\pi8.$$

Second integral. With $u=t^2$ the bounds become $0$ and $16$:
$$\int_0^4te^{t^2}\,dt=\frac12\int_0^{16}e^u\,du=\frac{e^{16}-1}2.$$

Remark. The second answer is huge ($\approx4.4\cdot10^6$) because $e^{t^2}$ explodes: always sanity-check the order of magnitude.