Question

$\int \frac{e^{(x^2+4\ln x)}-x^3{e}^{x^2}}{x-1}dx$ is equal to

• A. $\left(\frac{e^{2\ln x}-1}{2}\right)e^{x^2}+C$
• B. $\frac{(x-1)x{e}^{x^2}}{2}+C$
• C. $\frac{(x^2-1)}{2x}-e^{x^2}+C$
• D. none of these

Answer 1

The correct option is A $\left(\frac{e^{2\ln x}-1}{2}\right)e^{x^2}+C$

$\int \frac{e^{(x^2+4\ln x)}-x^3{e}^{x^2}}{x-1}$
$=\int \frac{e^{x^2}{x}^4-x^3{e}^{x^2}}{x-1}$
Let,
$x^2=t$
$2xdx=dt$
So,
$=\int \frac{e^t.x^4-x^3{e}^t}{x-1}dx$
$=\int \frac{e^t(x^4-x^3)}{x-1}dx$
$=\int e^t{x}^{3}dx$
$=\int e^t{x}^2.xdx$
$=\int \frac{e^t}{2}.tdt$

$=\frac{1}{2}(t.e^t-\int e^t+c)$

$=\frac{1}{2}\left(e^t\right(t-1\left)\right)+c$

$=\frac{1}{2}{e}^{x^2}(x^2-1)+c$