Answer
$\int x2^{x^{2}}dx=\frac{1}{2}\frac{2^{{x^{2}}}}{ln2}+constant$
Work Step by Step
Evaluate the integral $\int x2^{x^{2}}dx$.
Consider $x^{2}=t$
and
$\frac{d}{dx}x^{2}=dt$
$2xdx=dt$
$xdx=\frac{dt}{2}$
Now, $\int x2^{x^{2}}dx=\int 2^{t}\frac{dt}{2}$
$=\frac{1}{2}\frac{2^{t}}{ln2}+constant$
Hence, $\int x2^{x^{2}}dx=\frac{1}{2}\frac{2^{{x^{2}}}}{ln2}+constant$