`Ifint(x^4+1)/(x^6+1)dx=tan^(-1)f(x)-2/3tan^(-1)g(x)+C ,t h e n`
a) both `f(x)a n dg(x)` are odd functions
`b) f(x)` is monotonic function
`c) f(x)=g(x)` has no real roots
`d) int(f(x))/(g(x))dx=-1/x+3/(x^3)+c`

Options

A. both `f(x)` and `g(x)` are odd functions
B. `f(x)` is one-one function
C. `f(x)=g(x)` has no real roots
D. `int (f(x))/(g(x))dx=(1)/(x)+(3)/(x^(3))+c`

Answer

A, C, D

Explanation

Let `I=int((x^(4)+1))/((x^(6)+1))dt=int((x^(2)+1)^(2)-2x^(2))/((x^(2)+1)(x^(4)-x^(2)+1))dx`
`=int((x^(2)+1)dx)/((x^(4)-x^(2)+1))-2 int (x^(2)dx)/((x^(6)+1))`
`=int((1+(1)/(x^(2)))dx)/((x^(2)-1+(1)/(x^(2))))-2 int (x^(2)dx)/((x^(3))^(2)+1)`
In the first integral, put `x-(1)/(x)=t," i.e., " (1+(1)/(x^(2)))dx=dt`
and in the second integral put `x^(3) =u, " i.e., " x^(2)dx=(du)/(3)`
Then ` I=int (dt)/(1+t^(2))-(2)/(3)int(du)/(1+u^(2))=tan^(-1) t-(2)/(3)tan^(-1)u +C`
`=tan^(-1)(x-(1)/(x))-(2)/(3)tan^(-1)(x^(3))+C`
Here, `f(x)=x-(1)/(x) and g(x)=x^(3)`
Draw the graphs of `f(x) and g(x)` .
We find that `f(x)` is many-one and `f(x)=g(x)` has no real roots.
`int(f(x))/(g(x))dx=int(x-(1)/(x))/(x^(3))dx=int((1)/(x^(2))-(1)/(x^(4)))dx= -(1)/(x)+(3)/(x^(3))+C`