Answer
a) $u=3x^2-5$
$\hspace{0.4cm}du=6xdx$
b) $u=1-x$
$\hspace{0.4cm}du=-dx$
c) $u=2x^3+1$
$\hspace{0.4cm}du=6x^2dx$
d) $u=x^4$
$\hspace{0.4cm}du=4x^3dx$
Work Step by Step
a) $\int (3x^2-5)^4 2xdx$
Choose $u=3x^2-5$
$du=6xdx$
we can substitute $\frac{1}{3}du$ instead of $2xdx$.
Now the integral becomes $\int \frac{u^4}{3}du$
b) $\int \sqrt{1-x} dx$
Choose $u=1-x$
$du=-dx$
we can substitute $-du$ instead of $dx$.
Now the integral becomes $\int -\sqrt{u}du$
c)$\int \frac{x^2}{2x^3+1}dx$
Choose $u=2x^3+1$
$du=6x^2dx$
we can substitute $\frac{1}{6}du$ instead of $x^2dx$.
Now the integral becomes $\int \frac{1}{6u}du$
d)$\int 4x^3e^{x^4}dx$
Choose $u=x^4$
$du=4x^3dx$
we can substitute $du$ instead of $4x^3dx$.
Now the integral becomes $\int e^udu$
