∫(x)/(4x^2+8x+4) How to solve the Integration by parts

tnx

Given:

∫{(x)/(4x² + 8x + 4)}dx = ?

Factor out (1/4)

∫{(x)/(4x² + 8x + 4)}dx = (1/4)∫{(x)/(x² + 2x + 1)}dx

Notice that x² + 2x + 1 is of the form x² + 2ax + a² where a = 1. This means that it can be factored into a perfect square (x + 1)²:

∫{(x)/(4x² + 8x + 4)}dx = (1/4)∫{(x)/(x + 1)²}dx

Integrate by parts:

∫{u}dv = (u)(v) - ∫{v}du

let u = x, then du = dx
let dv = 1/(x + 1)², then v = ∫{1/(x + 1)²}dx = -1/(x + 1)

∫{(x)/(4x² + 8x + 4)}dx = (1/4)[-x/(x + 1) + ∫{1/(x + 1)}dx]

The last integral is just the natural log:

∫{(x)/(4x² + 8x + 4)}dx = (1/4)[-x/(x + 1) + ln|x + 1|] + C

TNX

Report Abuse

4x^2+8x+4 = 4x^2+4x+4x+4 = 4x(x+1)+4(x+1) = (x+1)(4x+4)=4(x+1)^2

∫x dx/(4x^2+8x+4) = (1/4) ∫x/(x+1)^2 dx

Integrate by parts --- just consider ∫x/(x+1)^2 dx and multiply by 1/4 at the end

dv=1/(x+1)^2 ; v=-1/(x+1)
u = x; du = 1;

∫ u dv = u v - ∫ v du
∫x/(x+1)^2 dx = -x/(x+1) - ∫ [-1/(x+1) dx = -x/(x+1) + ∫ [dx/(x+1)

= -x/(x+1) + ln(x+1)
multiply by 1/4

= -x/4(x+1) + ln(x+1)/4 + C