Question

Evaluate ∫ ln (8x) dx.

Answer 1

`\displaystyle{\int \ln(8x)\, dx}=x\ln(8x)-x+C`

Explanation:

So we have the integral:

`\displaystyle{\int \ln(8x)\, dx}`

First, let's do a substitution to simplify the inside of natural log. Let y equal to 8x. So:

`y=8x`

Differentiate with respect to x:

`dy=8\, dx`

Divide both sides by 8:

`(1)/(8)\, dy=dx`

Substitute this into our original integral. Therefore:

`\displaystyle{=\int (1)/(8)\ln(y)\, dy`

Move the co-efficient outside:

`\displaystyle{=(1)/(8)\int\ln(y)\, dy`

Now, perform integration by parts. Integration by parts is as follows:

`\displaystyle{\int u\, dv=uv-\int v\, du}`

Note that our integral is the same as saying:

`\displaystyle{=(1)/(8)\int 1\cdot\ln(y)\, dy`

Let u equal the natural log and let dv equal 1 dy. Therefore:

`u=\ln(y)`

Differentiate:

`du=(1)/(y)\, dy`

And let dv equal 1 dy. Thus:

`dv=1\, dy`

Integrate:

`v=y`

Perform integration by parts:

`\displaystyle{=(1)/(8)(y\ln(y)-\int y((1)/(y))\, dy}`

Simplify:

`\displaystyle{=(1)/(8)(y\ln(y)-\int 1\, dy)}`

Evaluate the integral:

`\displaystyle{=(1)/(8)(y\ln(y)-y)}`

Substitute 8x for y:

`\displaystyle{=(1)/(8)(8x\ln(8x)-8x)}`

Distribute:

`=x\ln(8x)-x`

Constant of Integration:

`=x\ln(8x)-x+C`

And we have our answer.

Edit: Typo