Integration by Parts with a logarithm Im not sure how i go about integrating a function being:
$$\int^{\sqrt{2}}_{1} r^{3}\log({r^2}) dr$$
I assume you use parts but it keeps making me go round in circles.
Any help would be appreciated.
My attempt was:
$$u=log(r^{2}) \rightarrow u'=\frac{2}{r},$$ $$v=\frac{r^4}{4} \rightarrow v'=r^{3}$$
so using the equation $uv- \int uv'$ i get:
$$\frac{r^{4}}{4}log(r^{2})- \int log(r^{2})r^{3}$$
 A: Make this substitution before you use Integration by parts:
$$u = r^2$$ 
A: Recall that $\log(r^2) = 2\log r$.  So then
\begin{align}
  \int_1^{\sqrt{2}} r^3 \log(r^2) \, dr
    &= \int_1^{\sqrt{2}} r^3 \cdot 2 \log r \, dr\\[0.3cm]
    &= 2\int_1^{\sqrt{2}} r^3 \log r \, dr
\end{align}
Now let $u = \log r$ so that $dv = r^3 \, dr$, and $du = \dfrac{1}{r} \, dr$ (I'm assuming by $\log$ you mean natural log and not common log) and $v = \dfrac{1}{4}r^4$.  So then
\begin{align}
  2\int_1^{\sqrt{2}} r^3 \log r \, dr
    &= 2\left[\frac{1}{4}r^4\log r\bigg|_1^{\sqrt{2}} - \int_1^{\sqrt{2}} \frac{1}{4}r^4 \cdot \frac{1}{r} \, dr\right]
\end{align}
Can you take it from here?
A: You are not working by parts the right way.
$$\int u'vdr=uv-\int uv'dr$$ yields
$$\int r^3\ln(r^2)dr=\frac{r^4}4\ln(r^2)-\int\frac{r^4}4\frac{2dr}r=\frac{r^4}4\ln(r^2)-\frac{r^4}8.$$
A: Logarithms are good to differentiate, powers are good to integrate.  So call this $\int u \; dv$ with $u = \log(r^3)$ and $dv = r^3\; dr$.
A: $$\int_{1}^{\sqrt{2}}r^3\ln(r^2)\space\text{d}r=$$

Integrate by parts, $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where:
$$f=\ln(r),\text{d}g=r^3\space\text{d}r,\text{d}f=\frac{1}{r}\space\text{d}r,g=\frac{r^4}{4}$$

$$\left[\frac{r^4\ln(r)}{2}\right]_{1}^{\sqrt{2}}-\frac{1}{2}\int_{1}^{\sqrt{2}}r^3\space\text{d}r=$$

Use $\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$:

$$\left[\frac{r^4\ln(r)}{2}\right]_{1}^{\sqrt{2}}-\frac{1}{2}\left[\frac{r^4}{4}\right]_{1}^{\sqrt{2}}=$$
$$\frac{1}{2}\left[r^4\ln(r)\right]_{1}^{\sqrt{2}}-\frac{1}{8}\left[r^4\right]_{1}^{\sqrt{2}}=$$
$$\frac{1}{2}\left(\left(\sqrt{2}\right)^4\ln(\sqrt{2})-1^4\ln(1)\right)-\frac{1}{8}\left(\left(\sqrt{2}\right)^4-1^4\right)=$$
$$\frac{1}{2}\left(4\cdot\ln(\sqrt{2})-0\right)-\frac{1}{8}\left(4-1\right)=$$
$$2\ln(\sqrt{2})-\frac{3}{8}=$$
$$\ln(2)-\frac{3}{8}$$
