JEE Mains Indefinite Integration 2

Let $\displaystyle I = \int {\left( \dfrac{\ln x -1}{1+ (\ln x)^2} \right)}^2 \,dx$

Substituting $\ln x = u \implies x = e^u \implies dx = e^u du$ , we get

$\begin{aligned} \displaystyle I &= \int {\left( \dfrac{u -1}{1+ u^2} \right)}^2 e^u \,du \\ \displaystyle &= \int \dfrac{u^2+1-2u}{(1+u^2)^2} e^u \,du \\ \displaystyle &= \int \left( \color{#20A900}{\dfrac{1}{1+u^2}} + \color{#D61F06}{\dfrac{(-2u)}{(1+u^2)^2}} \right) e^u \,du \\ \displaystyle &= \dfrac{e^u}{1+u^2} + C & \small{\text{Using } \displaystyle \int ( {\color{#20A900}{ f(x)}} +{ \color{#D61F06}{f'(x)}}) e^x \,dx = e^x f(x) + C} \\ &= \dfrac{x}{(\ln x)^2 +1} + C \\ \end{aligned}$

Easily visible from the options

Because differentiation of (lnx)^2 gives 1\x so x should be there in numerator