How Does Integration by Parts Differ in Solving $\int x^8 \ln x^9 \, dx$?

[karush]

$\Large{242.8.2.54} \\ $
solve by $u={x}^{8}$ then by IBP

$\displaystyle
I_{54}=\int{x^8}\ln{x^9} \, dx
=9\left(\dfrac{x^9\ln\left(x\right)}{9}-\dfrac{x^9}{81}\right) \\$
$u=x^8. \ \ du=8x^7 dx \ \ x=\sqrt[8]{u}$

how would IBP be any different?

 

[Greg]

I believe you'd want to use the substitution

$$t=x^9,\quad\dfrac{dt}{9}=x^8\,dx$$

so the integral becomes

$$\dfrac19\int\ln(t)\,dt$$

Now, to apply IBP, use

$$u=\ln(t),\,du=\dfrac1t\,dt;\quad dv=dt,\,v=t$$

Can you proceed from there?

 

[karush]

$$t=x^9,\quad\dfrac{dt}{9}=x^8\,dx \\

\dfrac19\int\ln(t)\,dt$$
IBP
$$\begin{align}
u&= \ln\left({t}\right) &dv&=dt\\
du&=\frac{1}{t}dt &v&=t&
\end{align}\\$$

$$\frac{1}{9}\left[uv-\int v \, du \right]
\implies
\frac{1}{9}\left[t \ln\left({t}\right)-\int t \frac{1}{t}dt\right]
\implies
\frac{1}{9} \left[t\ln\left({t}\right)-t\right]$$

back substitute $t=x^9$
$$\displaystyle
\frac{1}{9} \left[x^9\ln\left({x^9}\right)-x^9\right]+C$$

something ?

 

[Greg]

That's correct (and it's equivalent to the answer you gave in your initial post with the constant of integration added in).

 

[Prove It]

I would have used an entirely different substitution...

$\displaystyle \begin{align*} \int{ x^8\ln{ \left( x^9 \right) } \,\mathrm{d}x } &= \int{ 9\,x^8\ln{(x)} \,\mathrm{d}x } \\ &= \int{ \frac{9\,x^9\ln{(x)}}{x}\,\mathrm{d}x } \end{align*}$

then with the substitution $\displaystyle \begin{align*} u = \ln{(x)} \implies \mathrm{d}u = \frac{1}{x}\,\mathrm{d}x \end{align*}$ we have

$\displaystyle \begin{align*} \int{ \frac{9\,x^9\ln{(x)}}{x}\,\mathrm{d}x } &= \int{ 9\,u\,\mathrm{e}^{ 9\,u }\,\mathrm{d}u } \end{align*}$

and then apply Integration by Parts.

 

[soroban]

$\displaystyle I \;=\; \int{x^8}\ln{x^9} \, dx \;=\;x^9\ln x - \tfrac{1}{9}x^9 + C \\$

$$\text{We have: }\;I \;=\; 9\int x^8\ln x\,dx$$

$$\begin{array}{ccccccccc} u&=& \ln x && dv &=& x^8dx \\ du &=& \frac{dx}{x} && v &=& \frac{1}{9}x^9 \end{array}$$

$$I \;=\; 9\left[\tfrac{1}{9}x^9\ln x - \tfrac{1}{9}\int x^8dx\right] \;=\; 9\left[\tfrac{1}{9}x^9\ln x - \tfrac{1}{81}x^9\right] + C$$

$$I \;=\;x^9\ln x - \tfrac{1}{9}x^9 + C$$