Question

Solve :

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

• A. $\frac{x}{x^2+1}+c$
• B. $\frac{\ln x}{{\left(\ln x\right)}^2+1}+c$
• C. $\frac{x}{{\left(\ln x\right)}^2+1}+c$
• D. $e^x\left(\frac{x}{x^2+1}\right)+c$

Answer 1

The correct option is C $\frac{x}{{\left(\ln x\right)}^2+1}+c$

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

Put $\ln x=t\Rightarrow x=e^t\Rightarrow dx=e^{t}dt$

$\therefore I=\int e^t{\left(\frac{t-1}{t^2+1}\right)}^{2}dt=\int e^t(\frac{1}{t^2+1}-\frac{2t}{{(t^2+1)}^2})dt$

we know that $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^x\cdot f\left(x\right)+cwheref\left(x\right)=\frac{1}{t^2+1},f^{\prime }\left(x\right)=\frac{-2t}{(t^2+1{)}^2}$

$\therefore I=\frac{e^t}{t^2+1}+c=\frac{x}{{\left(\ln x\right)}^2+1}+c$