Question

Let $\int \frac{x^4+1}{x^6+1}dx={\tan }^{-1}(f(x))-\frac{2}{3}{\tan }^{-1}(g(x))+C$, where $C$ is constant of integration. where $f(1)=0$ and $g(0)=0$.
Number of roots of the equation $g(x)-xf(x)=0$ is equal to

• A. 3 real and distinct roots
• B. 3 real and coincident roots
• C. one real and two imaginary roots
• D. 3 real roots of which two are coincident

Answer 1

Answer: (C)

$I=\int \frac{x^4+1}{x^6+1}dx=\int \frac{{\left(x^2+1\right)}^2-2x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}dx=\int \frac{x^2+1}{x^4-x^2+1}dx-2\int \frac{x^{2}dx}{x^6+1}$

$=\int \frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}dx-2\int \frac{x^{2}dx}{{\left(x^3\right)}^2+1}$
$={\tan }^{-1}\left(x-\frac{1}{x}\right)-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+c$
$\Rightarrow f(x)=x-\frac{1}{x}\text{ and }g(x)=x^3$
hence $g(x)-xf(x)=0$ gives $F(x)=x^3-x^2+1$ and $D>0$ for $F^{\prime }(x)=0$
$\because F(0)F\left(\frac{2}{3}\right)>0\Rightarrow$ therefore only one real roots of $F(x)=0$
$\therefore$ All the answer the obvious.