Question:
Write a value of $\int \mathrm{e}^{2 \mathrm{x}^{2}+\ln \mathrm{x}} \mathrm{dx}$
Solution:
We know that $\mathrm{e}^{\mathrm{a}+\mathrm{b}}=\mathrm{e}^{\mathrm{a}} \mathrm{e}^{\mathrm{b}}$
$y=\int e^{2 x^{2}} e^{\ln x} d x$
$y=\int e^{2 x^{2}} x d x$
Let, $x^{2}=t$
Differentiating both sides with respect to $x$
$\frac{d t}{d x}=2 x$
$\Rightarrow \frac{1}{2} d t=x d x$
$y=\int \frac{1}{2} e^{2 t} d t$
Use formula $\int e^{a+b t}=\frac{e^{a+b t}}{b}$
$y=\frac{1}{2} \frac{e^{2 t}}{2}+c$
Again, put $\mathrm{t}=\mathrm{x}^{2}$
$y=\frac{e^{2 x^{2}}}{4}+c$