Question

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

Answer 1

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