Evaluate the following integrals

Question

∫3x/(1+x⁴)dx

I separated the two equations into 3x(1/(1+x^4) and tried u substitution but neither 3x nor 1+x^4 really worked so I wasn't sure how to do this equation.

∫1/(√3-2x-x²)dx [Hint: complete the square]

Answer 1

∫ [1 / √(-(x²+2x-3))]dx = ∫ [1 / √(-(x²+2x+1-4))]dx = ∫ [1 / √(4 - (x+1)²)dx = ∫ [ 1 / (√(4(1 - (x+1)²/4)) dx]

= (1/2) ∫ [1 / √(1 - ((x+1)/2)²)]dx

Let u = (x+1)/2. Then du = (1/2)dx

So, we have ∫ [1 / √(1 - u²)] du = Arcsin u + C = Arcsin [ (x+1)/2 ] + C

Answer 2

f(x)= 3x/(1+x^4)

let u=x^2, du=2xdx, dx=du/2x

F(u) = (3/2) integral of 1/(u^2+1) du

use an integral table

F(u) = (3/2)arctan(u) + c

substitute back u=x^2

= (3/2)arctan(x^2)+ c

derivative of arctan(u) = 1/(u^2+1)