MHB Integration by Substitution: Solving Integrals with u-Substitution

[karush]

$$\int\frac{3x-1}{\left(3{x}^{2}-2x+1\right)^4} dx$$

$$u=\left(3{x}^{2}-2x+1\right) du=6x-2\ dx=2\left(3x-1\right)dx$$

got this far but when I tried to complete the ans was wrong
the correct ans is
$\frac{-1}{6\left(3x^2-2x+1 \right)^3}+C$

 

[mathmari]

$$\int\frac{3x-1}{\left(3{x}^{2}-2x+1\right)^4} dx$$

$$u=\left(3{x}^{2}-2x+1\right) du=6x-2\ dx=2\left(3x-1\right)dx$$

got this far but when I tried to complete the ans was wrong
the correct ans is
$\frac{-1}{6\left(3x^2-2x+1 \right)^3}+C$

$$u=3{x}^{2}-2x+1 \Rightarrow du=6x-2\ dx=2\left(3x-1\right)dx \Rightarrow \frac{1}{2}du=\left(3x-1\right)dx$$

$$\int\frac{3x-1}{\left(3{x}^{2}-2x+1\right)^4} dx=\int \frac{1}{2u^4}du=\frac{1}{2} \int u^{-4}du=\frac{1}{2}\frac{1}{-3}u^{-3} +C =\frac{-1}{6u^3}+C =\frac{-1}{6\left(3{x}^{2}-2x+1\right)^3}+C$$

 

[karush]

I see I didn't have the $\frac{1}{2} du $ I had $2du$