Question

The value of the integral $\int \frac{e^{x^2+4\ln x}-x^3{e}^{x^2}}{x-1}dx$ equal to:

• A. $\left(\frac{e^{3\ln x}-e^{\ln x}}{2x}\right)e^{x^2}+C$
• 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 $\left(\frac{e^{3\ln x}-e^{\ln x}}{2x}\right)e^{x^2}+C$

Let $I=\int \frac{e^{x^2+4\ln x}-x^3{e}^{x^2}}{x-1}dx$
$=\int \frac{e^{x^2}{e}^{\ln x^4}-x^3{e}^{x^2}}{x-1}dx$
$=\int \frac{e^{x^2}{x}^3(x-1)}{x-1}dx[\because e^{\ln x}=x]$
$\Rightarrow I=\int x^3{e}^{x^2}dx=\frac{1}{2}\int t{e}^{t}dt,$
$($where $t=x^2\Rightarrow dt=2xdx)$
$=\frac{1}{2}[t{e}^t-\int e^{t}dt]$
$\Rightarrow I=\frac{1}{2}(t-1)e^t+C$
$=\frac{1}{2}(x^2-1)e^{x^2}+C$