Question

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

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

Answer 1

The correct option is A

$\frac{x}{\left[{\left(\log x\right)}^2+1\right]}+c$

Explanation for the correct option:

Find the value of the given integration:

Consider the given equation as,

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

Let us assume that,

$$\begin{gathered} \log x=t\text{⇒}x=e^t \\ \frac{1}{x}dx=dt \\ \Rightarrow dx=xdt \\ \Rightarrow dx=e^{t}dt \end{gathered}$$[differentiating]

Then $I$ becomes,

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

Consider,

$$\begin{gathered} f\left(t\right)=\frac{1}{1+t^2} \\ f\text{'}\left(t\right)=\frac{-2t}{{\left(1+t^2\right)}^2} \end{gathered}$$

Then $I$ becomes,

$$\begin{gathered} I=\int e^t\left[f\left(t\right)+f\text{'}\left(t\right)\right]dt\left(\text{where},\int e^t\left[f\left(t\right)+f\text{'}\left(t\right)\right]dt=f\left(x\right)+c\right) \\ =e^t.f\left(t\right)+c \\ =e^t\frac{1}{1+t^2}+c \end{gathered}$$

Substitute the value of $t$ in the above Equation

$$\begin{gathered} I=e^{\log x}\left(\frac{1}{1+{\left(\log x\right)}^2}\right)+c \\ =x\left(\frac{1}{1+{\left(\log x\right)}^2}\right)+c \\ I=\left(\frac{x}{1+{\left(\log x\right)}^2}\right)+c \end{gathered}$$

Hence, the correct answer is Option (A).