Question

$\int {\left(\frac{\left(\ln x\right)-1}{1+(\ln x{)}^2}\right)}^{2}dx$ is equal to
(where $c$ is constant of integration)

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

Answer 1

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

Let $\ln x=t$
$\Rightarrow x=e^t$
$\Rightarrow dx=e^{t}dt$
$\therefore \int {\left(\frac{t-1}{1+t^2}\right)}^2{e}^{t}dt=\int e^t\left(\frac{t^2-2t+1}{{(1+t^2)}^2}\right)dt=\int e^t(\frac{t^2+1}{{(1+t^2)}^2}+\frac{(-2t)}{{(1+t^2)}^2})dt=\int e^t(\frac{1}{1+t^2}+\frac{-2t}{(1+t^2{)}^2})dt\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }=\frac{e^t}{1+t^2}+c[\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+c]$
Hence, $\int {\left(\frac{\left(\ln x\right)-1}{1+(\ln x{)}^2}\right)}^{2}dx=\frac{x}{1+{\left(\ln x\right)}^2}+c$