Question

Solve$\int {\left\{\frac{(logx-1}{1+(logx{)}^2}\right\}}^{2}dx=$

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

Answer 1

The correct option is D $\frac{x}{(logx{)}^2+1}+c$

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

Let $logx=t$

$\therefore \frac{1}{x}dx=dt\Rightarrow dx=e^{t}dt$

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

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

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

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

We know, $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$

In $I$, $f\left(t\right)=\frac{1}{(1+t^2)}$ and $f^{\prime }\left(t\right)=\frac{2t}{(1+t^2{)}^2}$

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

i.e $I=x\frac{1}{1+(logx{)}^2}+c$