Integral how to choose the good substitution? I have a problem with this integral. I don't know what kind of substitution to use.

$$\int t\sqrt{1+9t^4}dt$$
$$
\Big( t\,dt\Big)
$$
What you see above suggests at least considering $u=t^2$ so that $du=2t\,dt$, and then what you see above is $\frac 1 2 \, du$.

Since you have $9t^4 = (3t^2)^2$, using $u=3t^2$ works; then $du = 6t\,dt$ and $\frac 1 6 \, du = t\,dt$.

And then you have $9t^4 = (3t^2)^2 = u^2$, so $\displaystyle\int \sqrt{1+9t^4}\ t\, dt = \frac 1 6 \int\sqrt{1+u^2}\  du$.
$3t^2 = \sin x$, $t\,dt=\frac{1}{6} \cos x \,dx$. So, we have,

$$\int \cos x (1/6) \cos x \, dx =\int \frac{1}{12}(1+\cos2x) \, x 
= \frac {1} {12} (x + \sin x \cos x)+c=\frac{1}{12}\left(\arcsin(3t^2)+3t^2\sqrt{1-9t^4}\right)+c.$$