Question

$\int \frac{x^2-1}{(x^4+3x^2+1){\tan }^{-1}\left(x+{\displaystyle \frac{1}{x}}\right)}dx=$

• A. ${\tan }^{-1}\left(x+\frac{1}{x}\right)+c$
• B. $co{t}^{-1}\left(x+\frac{1}{x}\right)+c$
• C. $\log \left(x+\frac{1}{x}\right)+c$
• D. $\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)\right)+c$

Answer 1

The correct option is D

$\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)\right)+c$

Explanation for the correct answer.

Evaluate the given integral.

An integral $I=\int \frac{x^2-1}{(x^4+3x^2+1){\tan }^{-1}\left(x+{\displaystyle \frac{1}{x}}\right)}dx$ is given.

Rewrite the given integral as follows:

$$\begin{gathered} I=\int \frac{1-{\displaystyle \frac{1}{x^2}}}{\left(x^2+3+{\displaystyle \frac{1}{x^2}}\right){\tan }^{-1}\left(x+{\displaystyle \frac{1}{x}}\right)}dx \\ \Rightarrow I=\int \frac{1-{\displaystyle \frac{1}{x^2}}}{\left({\left(x+{\displaystyle \frac{1}{x}}\right)}^2+1\right){\tan }^{-1}\left(x+{\displaystyle \frac{1}{x}}\right)}dx \end{gathered}$$

Now, assume that, $x+\frac{1}{x}=u$

Differentiate both sides with respect to $x$.

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

Write the integral in terms of $u$ by substituting $x+\frac{1}{x}=u$.

$I=\int \frac{1}{\left(u^2+1\right){\tan }^{-1}\left(u\right)}du$

Again, Assume that ${\tan }^{-1}u=v$.

Differentiate both sides with respect to $u$.

$\frac{1}{u^2+1}du=dv$

Write the integral in terms of $v$ by substituting ${\tan }^{-1}u=v$.

$$\begin{gathered} I=\int \frac{1}{v}dv \\ \Rightarrow I=\log \left(v\right)+c \end{gathered}$$

Substitute back ${\tan }^{-1}u=v$.

$I=\log \left({\tan }^{-1}u\right)+c$

Substitute back $x+\frac{1}{x}=u$.

$I=\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)\right)+c$

Therefore, after evaluating the given integral we get $\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)\right)+c$.

Hence, option $D$ is the correct answer.