Answer:
[tex]\displaystyle{\int \ln(8x)\, dx}=x\ln(8x)-x+C[/tex]
Step-by-step explanation:
So we have the integral:
[tex]\displaystyle{\int \ln(8x)\, dx}[/tex]
First, let's do a substitution to simplify the inside of natural log. Let y equal to 8x. So:
[tex]y=8x[/tex]
Differentiate with respect to x:
[tex]dy=8\, dx[/tex]
Divide both sides by 8:
[tex]\frac{1}{8}\, dy=dx[/tex]
Substitute this into our original integral. Therefore:
[tex]\displaystyle{=\int \frac{1}{8}\ln(y)\, dy[/tex]
Move the co-efficient outside:
[tex]\displaystyle{=\frac{1}{8}\int\ln(y)\, dy[/tex]
Now, perform integration by parts. Integration by parts is as follows:
[tex]\displaystyle{\int u\, dv=uv-\int v\, du}[/tex]
Note that our integral is the same as saying:
[tex]\displaystyle{=\frac{1}{8}\int 1\cdot\ln(y)\, dy[/tex]
Let u equal the natural log and let dv equal 1 dy. Therefore:
[tex]u=\ln(y)[/tex]
Differentiate:
[tex]du=\frac{1}{y}\, dy[/tex]
And let dv equal 1 dy. Thus:
[tex]dv=1\, dy[/tex]
Integrate:
[tex]v=y[/tex]
Perform integration by parts:
[tex]\displaystyle{=\frac{1}{8}(y\ln(y)-\int y(\frac{1}{y})\, dy}[/tex]
Simplify:
[tex]\displaystyle{=\frac{1}{8}(y\ln(y)-\int 1\, dy)}[/tex]
Evaluate the integral:
[tex]\displaystyle{=\frac{1}{8}(y\ln(y)-y)}[/tex]
Substitute 8x for y:
[tex]\displaystyle{=\frac{1}{8}(8x\ln(8x)-8x)}[/tex]
Distribute:
[tex]=x\ln(8x)-x[/tex]
Constant of Integration:
[tex]=x\ln(8x)-x+C[/tex]
And we have our answer.
Edit: Typo
