Integrate dx / x(2x^3 +1)^2

show steps please :)

Set u = x^3. Then du = 3x^2 dx. Then dx = du/3x^2.

Then we get du/(3u(2u+1)^2) as the integrand.

We then use partial fractions to solve this. Solving the linear equations gives:

(1/u - 2/(2u+1) - 2/(2u+1)^2)*du/3

Integrating all of these individually gives (ln|u|-ln|2u+1| + 1/(2u+1) + C)/3

Then substituting u = x^3 back gives

(ln|x^3|-ln|2x^3+1| + 1/(2x^3+1) )/3+ C

Or

ln|x| - ln|2x^3+1|/3 + 1/(6x^3+3) + C

EDIT: Sorry, forgot to distribute in the fractions

∫1 /[x(2x³ +1)²] dx=
=(1/6)∫6x² /[x³(2x³ +1)²] dx=
=(1/6)∫1/[x³(2x³ +1)²] d(2x³ +1)=
Let v=2x³ +1, x³=½(v-1)
=(1/6)∫2/[(v-1)v²] dv=
=(1/3)∫ 1/[(v-1)v²] dv

1/[(v-1)v²]=a/v + b/v² + с/(v-1)
1=av(v-1) + b(v-1) + сv²
1=(a+с)v²+ (b-a)v -b

System
0=a+с
0=b-a
1=-b

a=-1, b=-1 and c=1
hence
1/[(v-1)v²]=-1/v - 1/v² + 1/(v-1)

(1/3)∫ 1/[(v-1)v²] dv=(1/3)∫ -1/v - 1/v² + 1/(v-1) dv=
= -(1/3)ln|v| + (1/3)1/v + (1/3)ln|v-1|+C=
= -(1/3)ln|2x³ +1| + (1/3)1/(2x³ +1) + (1/3)ln|2x³ +1-1|+C=
= -(1/3)ln|2x³ +1| + (1/3)1/(2x³ +1) + (1/3)ln|2x³|+C
http://www.wolframalpha.com/input/?i=%E2…

let 2x^3+1=z
then x^3=(z-1)/2 and 3x^2dx=dz
now, 2dz/3(z-1)(z)^2
let 1/(z^2 (z-1))=(Az+B)/z^2+C/(z-1)
=> 1=Az^2-Az+Bz-B+Cz^2
=> A+C=0, A-B=0, B=-1
=> A=-1, B=-1, C=1
Therefore, 2/3*dz/(z^2 (z-1))
2/3*[-ln z+1/z^2+ln (z-1)]+const.
put the value of z and get your desired answer.