Question

The value of the integral $\int \frac{e^{x^2+4\ln x}-x^3{e}^{x^2}}{x-1}dx$ equal to
(where $C$ is the constant of integration)

• A. $\frac{(x^2-1)e^{x^2}}{2}+C$
• B. $\frac{(x-1)x{e}^{x^2}}{2}+C$
• C. $\frac{(x^2-1)e^{x^2}}{2x}+C$
• D. $\frac{(x+1)e^{x^2}}{2x}+C$

Answer 1

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

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