∫(x)/(4x^2+8x+4) plz help me solve this integral

plz help me???

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

Let x + 1 = u:

u = x + 1
⇒ x = u − 1
⇒ dx/du = 1
⇒ dx = du

Substitute x + 1 = u, x = u − 1 and dx = du:

1/4 ∫ (u − 1) / u² du
= 1/4 ∫ u / u² − 1 / u² du
= 1/4 ∫ 1 / u − 1 / u² du
= 1/4 (ln(u) + 1 / u + C)

Back-substitute u = x + 1:

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

∫ (u − 1) / u² du=>∫ u / u² − 1 / u² du
How?

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(u-1)/u^2 = (u/u^2-1/u^2) hence above line

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∫(x)/(4x^2+8x+4)=(1/4)*∫(x)/(x^2+2x+1)
=(1/4)*∫(1/2)*(2*(x)+2)/(x^2+2x+1) - (1/4)*∫1/(x^2+2x+1)
=(1/4)*(1/2) ln(x^2+2x+1) +(1/4)* (x+1)^-1 +c

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

u = x + 1
du = dx
x = u - 1

(u - 1) * du / (4 * u^2) =>
(1/4) * (u * du / u^2 - 1 * du / u^2) =>
(1/4) * (du / u - du / u^2) =>
(1/4) * (ln|u| + 1/u) + C =>
(1/4) * (ln|x + 1| + 1 / (x + 1)) + C

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

See?