Question

Let $I=\underset{x_1}{\overset{x_2}{\int }}\frac{x^4-1}{x+x^5}dx,$ where $x_1$ and $x_2$ are positive real numbers satisfying $\frac{x_1^4+1}{x_2^4+1}=\frac{x_1}{x_2^2}.$ Then the value of $2I$ is

• A. $\ln \left(x_1\right)$
• B. $\ln \left(x_1^2\right)$
• C. $\ln \left(x_2\right)$
• D. $\frac{1}{2}\ln \left(x_2\right)$

Answer 1

The correct option is A $\ln \left(x_1\right)$

$I=\underset{x_1}{\overset{x_2}{\int }}\frac{x^4-1}{x+x^5}dx$
Divide numerator and denominator by $x^3$
$I=\underset{x_1}{\overset{x_2}{\int }}\frac{x-\frac{1}{x^3}}{\frac{1}{x^2}+x^2}dx$
Let $\frac{1}{x^2}+x^2=t$
$\Rightarrow 2x-\frac{2}{x^3}=\frac{dt}{dx}$

$\Rightarrow I=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\ln t$

$=\frac{1}{2}{\left[\ln (\frac{1}{x^2}+x^2)\right]}_{x_1}^{x_2}$

$=\frac{1}{2}\ln (\frac{x_2^4+1}{x_2^2}\times \frac{x_1^2}{x_1^4+1})$
$=\frac{1}{2}\ln \left(\frac{x_1^2}{x_1}\right)$

$\Rightarrow I=\frac{1}{2}\ln \left(x_1\right)$

$\Rightarrow 2I=\ln \left(x_1\right)$