How do i get to the integral of...?

Question

integral of x/(1+4x^2)

and integral of x/(sqrt(1-x^2))

thanks!!

srry im just really lost on tonights calc bc hwk

Answer 1

∫x/(1+4x²) dx =

Use u-substitution.  Let u = 1 + 4x².  Then du = 8x dx.  Since the integral above has x dx but not 8x dx, multiply the inside by 8 and the outside by 1/8.  Then the integral above becomes

(1/8)∫(8x dx)/(1+4x²) =

(1/8)∫du /(u) =

(1/8)∫du/u =

(1/8)ln(|u|) + C =

Replace u with 1 + 4x² from the u-substitution.

(1/8)ln(|1 + 4x²|) + C =

Since x² ≥ 0 then 1 + 4x² ≥ 1, which means that |1 + 4x²| = 1 + 4x²

(1/8)ln(1 + 4x²) + C

For the second problem,

∫x/√(1-x²) dx =

Use u-substitution with u = 1 - x².  Then du = -2x dx.  Since the integral above has x dx instead of -2x dx multiply the inside by -2 and the outside by -½.  Then the integral above becomes:

(-½)∫(-2x dx) / √(1-x²) =

(-½)∫du / √u =

(-½)∫u^(-½) du =

(-½)(2u^(½) + C) =

(-½)(2√u + C) =

(-½)(2√u) + C =

-√u + C =

C - √u =

Replace u with 1 - x² from the u-substitution.

C - √(1 - x²)

Hope this helps you!