Question

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

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

$=\int \frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}} d x-2 \int \frac{x^2 d x}{\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 $g(x)-x f(x)=0$ 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.