Question

Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration.

∫(8x+10) ln (5x) dx.

Answer 1

`\text{ln}(5x)(4x^2+10x)-2x^(2)-10x+C`

Explanation:

We have been given an definite integral
`\int \left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx`. We are asked to find the integral using integration by parts.

We will use Integration by parts formula to solve our given problem.

`\int\ vdv=uv-\int\ vdu`

Let
`u=\text{ln}(5x)` and
`v'=8x+10`.

Now, we need to find du and v using these values as shown below:

`(du)/(dx)=(du)/(dx)(\text{ln}(5x))`

Using chain rule, we will get:

`(du)/(dx)=(1)/(5x)*5`

`(du)/(dx)=(1)/(x)`

`du=(1)/(x)dx`

`v'=8x+10`

`v=(8x^(1+1))/(2)+10x`

`v=(8x^(2))/(2)+10x`

`v=4x^(2)+10x`

Upon substituting these values in integration by parts formula, we will get:

`\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int\ (4x^2+10x)(1)/(x)dx`

`\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int\ (4x^2)/(x)+(10x)/(x)dx`

`\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int 4x+10dx`

`\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-((4x^(2))/(2)+10x)+C`

`\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-2x^(2)-10x+C`

Therefore, our required integral would be
`\text{ln}(5x)(4x^2+10x)-2x^(2)-10x+C`.