Question

solution for the integral of x/x^4+1 dx

Answer 1

Let u=x², then du/dx=2x, xdx=½du; xdx/(x⁴+1)=(½du)/(u²+1).

Let u=tanθ, du/dθ=sec²θ, u²+1=sec²θ, so (½du)/(u²+1)=(½sec²θdθ)/sec²θ=½dθ.

∫xdx/(x⁴+1)=½∫dθ=θ/2+C (where C is the integration constant)=

½tan^-1(u)+C=½tan^-1(x²)+C.

CHECK

Differentiate ½tan^-1(x²)+C:

Let y=½tan^-1(x²)+C, tan(2y-2C)=x², 1+tan²(2y-2C)=sec²(2y-2C)=1+x⁴

2sec²(2y-2C)dy/dx=2x, sec²(2y-2C)dy/dx=x, dy/dx=x/(1+x⁴) which is the original integrand.