# What is the integral int (x^2+1)(x^3+3x)^4 dx?

`int (x^2+1)(x^3+3x)^4 dx`

Use U-substitution to solve this.

`U=x^3+3x `

Take the derivative. Notice that there is a common factor of three!

`dU= 3x^2 +3 dx= 3(x^2+1) dx`

Divide by three on both sides, and then we will see that we can plug in `(x^2+1) dx`

`1/3 dU = (x^2+1) dx`

Now that we know and , substitute these back to the integral.

`int (x^2+1)(x^3+3x)^4 dx = int U^4 (1/3 dU)`

Pull out the constant one-third outside of the integral.

` ``1/3int U^4 dU`

Integrate by power rule.

`1/3 [U^5 /5] +C`

Substitute back and simplify.

`1/3 [(x^3+3x)^5 /5] +C = 1/15 (x^3+3x)^5 +C`

Do NOT forget to add a +C at the end since this is an indefinite integral, which means the integral doesn't have bounds.

The answer is: `1/15 (x^3+3x)^5 +C`

The integral `int (x^2+1)(x^3+3x)^4 dx` has to be determined.

Using substitution is the simplest way to arrive at the result.

Let `y = x^3+3x`

Taking the derivative of both the sides

`dy/dx = 3x^2+3 = 3*(x^2+1)`

`dy/3 = (x^2+1) dx`

Substituting in the integral `int (x^2+1)(x^3+3x)^4 dx` , we get

`int y^4/3 dy`

= `y^5/15`

As `y = x^3+3x` , the integral `int (x^1+1)(x^3+3x)^4 dx = (x^3+3x)^5/15 + C`

**The integral `int (x^1+1)(x^3+3x)^4 dx = (x^3+3x)^5/15 + C` **