u-substitution of definite integral

Question

$$\int_1^3 \frac{(\ln(v+1))^2}{(v+1)}dv$$

Here is what I have done:

$u = v+1$

$\frac{du}{dx}=1$

$du = dv$

$$\int_2^4 \frac{(\ln(u))^2}{(u)}du$$

Then

$z=\ln(u)$

$\frac{dz}{du}=\frac{1}{u}$

$dz = \frac{du}{u}$

$$\int_2^4 \frac{(z)^2}{(u)^2}dz$$

And here I was told I was wrong. As there should only be $z^2$ left in the equation but I cant tell what is the problem. Any hint? Thanks.

Answer 1

Hint

With the substitution $$z=\ln(u)$$ and

$$dz=\frac{du}{u}$$

the integral $$\int_2^4\frac{(\ln(u))^2}{u}du$$

becomes

$$\int_{\ln(2)}^{\ln(4)}z^2dz$$

Answer 2

$$\int \frac{(\log u)^2}{u}du = \int (\log u)^2 \color{red}{\frac{du}u} = \int z^2 \color{red}{dz}.$$