# How to solve $\int{4x^5\ln(5x)dx}$ using an integral table

I've been trying to solve this indefinite integral for some hours already. The problem is as follows:

a) Solve $$\int{4x^5\ln(5x)dx}$$ using integration by parts (IbP).

b) Verify your answer in (b) by solving the same integral, but with an integral table.

I already solved it using IbP, I got$$\frac{x^6(6\ln(5x)-1)}{9}+C.$$

My issues are with the second part of the problem. I can't seem to find an appropriate formula for this particular form. I've tried using $$\int{v^n\ln(v)dv}=v^{n+1}\left[\frac{\ln(v)}{n+1}-\frac{1}{\left(n+1\right)^2}\right] (i) ,$$ and multiplying $$x^5$$ by $$5/5$$, so I can make a variable substitution of $$v=5x$$.

I later realized this is not valid since $$5x^5$$ is not the same as $$\left(5x\right)^5$$, and that conflicts with the statement $$v=5x$$; nonetheless, I still believe it should be totally possible to transform one of the two expressions ($$x^5$$ or $$\ln(5x)$$) so you can use (i).

I would be really thankful if someone could point out any mistakes I made on my thought process, or if there's another way to solve (b). Thanks!

If

$$I=\int 4x^5\ln(5x)~\mathrm{d}x$$

and we make the substitution

$$u=5x,\quad\mathrm{d}u=5~\mathrm{d}x,$$

then

$$I=\int\frac{4}{5}\left(\frac{u}{5}\right)^5\ln u~\mathrm{d}u=\frac{4}{5^6}\int u^5\ln u~\mathrm{d}u.$$

In this form, you can use the proposed formula from your table.

• Tysm, I didn't realize my mistake of not solving for x to make the correct substitution. I was getting 5u^5 instead of (u/5)^5. A newbie mistake cost me 2 hours and almost a headache! Oct 15, 2022 at 23:31

Another possible approach: \begin{align*} \int 4x^{5}\ln(5x)\mathrm{d}x & = \int4x^{5}(\ln(5) + \ln(x))\mathrm{d}x\\\\ & = 4\ln(5)\int x^{5}\mathrm{d}x + 4\int x^{5}\ln(x)\mathrm{d}x\\\\ & = \frac{2\ln(5)x^{6}}{3} + \frac{2x^{6}\ln(x)}{3} - \frac{2}{3}\int x^{5}\mathrm{d}x\\\\ & = \frac{2\ln(5)x^{6}}{3} + \frac{2x^{6}\ln(x)}{3} - \frac{x^{6}}{9}\\\\ & = \frac{2x^{6}(\ln(5) + \ln(x))}{3} - \frac{x^{6}}{9}\\\\ & = \frac{(6\ln(5x) - 1)x^{6}}{9} \end{align*}

\begin{aligned} \int 4 x^5 \ln (5 x) d x & \stackrel{IBP}{=} \frac{2}{3} \int \ln (5 x) d\left(x^6\right) \\ &=\frac{2}{3} x^6 \ln (5 x)-\frac{2}{3} \int x^4 \cdot \frac{1}{x} d x \\ &=\frac{2}{3} x^6 \ln (5 x)-\frac{x^6}{9}+C \end{aligned}