Question

Use integration by parts to find the integrals in Exercise.
x^3 ln x dx.

Answer 1

`\frac{\text{ln}(x)x^4}{4}-(x^4)/(16)+C`.

Explanation:

We have been given an indefinite integral
`\int \:x^3\:ln\:x\:dx`. We are asked to find the value of the integral using integration by parts.

`\int\: u\text{dv}=uv-\int\: v\text{du}`

Let
`u=\text{ln}(x)`,
`v'=x^3`.

Now, we will find du and v as shown below:

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

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

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

`v=(x^(3+1))/(3+1)=(x^(4))/(4)`

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

`\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*(x^4)/(4)-\int\: (x^4)/(4)*(1)/(x)dx`

`\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: (x^3)/(4)dx`

`\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-(1)/(4)\int\: x^3dx`

`\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-(1)/(4)*(x^(3+1))/(3+1)+C`

`\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-(1)/(4)*(x^4)/(4)+C`

`\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-(x^4)/(16)+C`

Therefore, our required integral would be
`\frac{\text{ln}(x)x^4}{4}-(x^4)/(16)+C`.