Question

If $\frac{\ln \left(e{x}^{x+1}\right)+{\left(\ln \left(x^{\sqrt{x}}\right)\right)}^2}{1+(x\ln x)\left(\ln \left(e^2{x}^x\right)\right)}dx=f(x)+C$, where $f(1)=0$ then $e^{\left(e^{f(2)}-1\right)}$ is equal to

Answer 1

Answer: 4

$I=\int \frac{1+\ln x+x\ln x+x{\ln }^{2}x}{1+x\ln x(2+x\ln x)}dx$
$=\int \frac{(1+x\ln x)(1+\ln x)}{(1+x\ln x{)}^2}dx=\int \frac{1+\ln x}{(1+x\ln x)}dx$
$=\ln (1+x\ln x)+C$
$\left\{\because (1+x\ln x{)}^{\prime }=1+\ln x\right\}$
$\Rightarrow f(x)=\ln (1+x\ln x)$
$\Rightarrow e^{f(2)}=1+2\ln 2$
$\Rightarrow e^{f(2)}-1=\ln 4$