Verified answer

(1) ∫ (1 - 2x)² dx / x

= ∫ (1 - 2x)(1 - 2x) dx / x

= ∫ (1 - 4x + 4x²) dx / x

= ∫ [(1/x) - (4x/x) + (4x²/x)] dx

= ∫ [(1/x) - 4 + 4x] dx

= ln |x| - 4x + 2x² + C

(2) ∫ (u² + 1)² du / u³

= ∫ (u² + 1)(u² + 1) du / u³

= ∫ (u^4 + 2u² + 1) du / u³

= ∫ [(u^4/u³) + (2u²/u³) + 1/u³] du

= ∫ [u + (2/u) + 1/u³] du

= (1/2)u² + 2 ln |u| - 1/(2u²) + C

(3) ∫ y dy / (1 + y²)^4

Let u = (1 + y²).

du/dy = 2y

du = 2y dy

(1/2) du = (1/2) 2y dy

(1/2) du = y dy

So, integral becomes

= ∫ (1/2) du / (u)^4

= (1/2) ∫ u^(-4) du

= (1/2) (-1/3)u‾³ + C

= -1/(6u³) + C

= -1/[6(1 + y²)³] + C

In response to you previous question:

Thank you sir, I hope you could teach me how you get the dy=(1/3)dz.

z = 3y - 4

dz/dy = 3

dz = 3 dy

Divide both sides by 3.

dz/3 = (3 dy)/3

(1/3) dz = dy

the first problem can be separated into 3 integrals

(1-2x)²/x = (1-2*x)*(1-2*x)/x = (4x^2-4x+1)/x = 4x-4+1/x

the integrals become

int(4*x)dx - int(4)dx + int(1/x)dx = 2x^2 - 4x + ln(x) +c

the second problem got cut off, can't help with it