Question

The value of integration $\int x^{(\ln x{)}^{\ln \left(\ln x\right)}}\frac{(\ln x{)}^{\ln \left(\ln x\right)}}{x}\left(2\ln \right(\ln x)+1)dx$ equals to

• A. ${(\ln x{)}^{\ln \left(\ln x\right)}}^2+c$
• B. $x^{\left(\ln x\right)}+c$
• C. $x^{(\ln x{)}^{\ln \left(\ln x\right)}}+c$
• D. $(\ln x{)}^{\ln \left(\ln x\right)}+c$

Answer 1

The correct option is C $x^{(\ln x{)}^{\ln \left(\ln x\right)}}+c$

Given :
$\int x^{(\ln x{)}^{\ln \left(\ln x\right)}}\frac{(\ln x{)}^{\ln \left(\ln x\right)}}{x}\left(2\ln \right(\ln x)+1)dx$
Let $t=x^{(\ln x{)}^{\ln \left(\ln x\right)}}\cdots \left(i\right)$
Taking $\ln$ both sides, we get
$\Rightarrow \ln t=\left(\ln x\right)(\ln x{)}^{\ln \left(\ln x\right)}\cdots (ii)$
Again, taking $\ln$ of both sides, we get
$\Rightarrow \ln \left(\ln t\right)=\ln \left(\ln x\right)+\ln \left(\ln x\right)\ln \left(\ln x\right)$
$\Rightarrow \ln \left(\ln t\right)=\ln \left(\ln x\right)+\left(\ln \right(\ln x){)}^2$

Differentiating w.r.t. $x$ we get
$\frac{1}{\ln t}\cdot \frac{1}{t}\frac{dt}{dx}=\frac{1}{x\ln x}+\frac{2\ln \left(\ln x\right)}{\ln x}\frac{1}{x}$

$\Rightarrow \frac{1}{\ln t}\cdot \frac{1}{t}\frac{dt}{dx}=\frac{2\ln \left(\ln x\right)+1}{x\ln x}$

$\Rightarrow \frac{dt}{dx}=\frac{t}{x}\cdot \frac{\ln t}{\ln x}\left(2\ln \right(\ln x)+1)$

from $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow \frac{dt}{dx}=x^{(\ln x{)}^{\ln \left(\ln x\right)}}\frac{(\ln x{)}^{\ln \left(\ln x\right)}}{x}\left(2\ln \right(\ln x)+1)$

$\therefore dt=x^{(\ln x{)}^{\ln \left(\ln x\right)}}\frac{(\ln x{)}^{\ln \left(\ln x\right)}}{x}\left(2\ln \right(\ln x)+1)dx$
Now,

$\int x^{(\ln x{)}^{\ln \left(\ln x\right)}}\frac{(\ln x{)}^{\ln \left(\ln x\right)}}{x}\left(2\ln \right(\ln x)+1)dx$

$=\int 1dt$

$=t+c$

$=x^{(\ln x{)}^{\ln \left(\ln x\right)}}+c$