The integral `int x^3*sqrt(x^2 - 25) dx` has to be determined.

`int x^3*sqrt(x^2 - 25) dx`

= `int x^2*x*sqrt(x^2 - 25) dx`

Let `x^2 - 25 = y`

`x^2 = y + 25`

`dy/dx = 2x`

`dy/2 = x*dx`

The integral `int x^3*sqrt(x^2 - 25) dx` can be written as:

...

## Check Out

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The integral `int x^3*sqrt(x^2 - 25) dx` has to be determined.

`int x^3*sqrt(x^2 - 25) dx`

= `int x^2*x*sqrt(x^2 - 25) dx`

Let `x^2 - 25 = y`

`x^2 = y + 25`

`dy/dx = 2x`

`dy/2 = x*dx`

The integral `int x^3*sqrt(x^2 - 25) dx` can be written as:

`int (y + 25)*sqrt y*(1/2)*dy`

= `(1/2)*int y*sqrt y + 25*sqrt y dy`

= `(1/2)*int y^(3/2) dy + (25/2)*int y^(1/2) dy`

= `(1/2)*(y^(5/2))/(5/2) + (25/2)*y^(3/2)/(3/2) + C`

= `(1/5)*(y^(5/2)) + (25/3)*y^(3/2) + C`

Substitute `y = x^2 - 25`

= `(1/5)*(x^2 - 25)^(5/2) + (25/3)*(x^2 - 25)^(3/2) + C`

The integral `int x^3*sqrt(x^2 - 25) dx` is `(1/5)*(x^2 - 25)^(5/2) + (25/3)*(x^2 - 25)^(3/2) + C`