Question

Evaluate this integral?

`int_0^ln10 (e^(x)sqrt(e^(x)-1))/(e^(x)+8)dx`

Answer 1

`int_0^ln10 \ ( e^x sqrt( e^x-1 ) ) / ( e^x+8 ) \ dx = 6 - (3pi)/2`

Explanation:

We seek:

`I = int_0^ln10 \ ( e^x sqrt( e^x-1 ) ) / ( e^x+8 ) \ dx`

Let us attempt a substitution of the form:

`u = sqrt(e^x-1) => u^2 = e^x - 1`

Differentiating wrt `x`, we have:

`2u(du)/dx = e^x => 2u (du)/dx sqrt(e^x-1)= e^xsqrt(e^x-1)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ => 2u^2 (du)/dx = e^xsqrt(e^x-1)`

We must also change the limits if integration:

When `x={ (ln10), (0) :} => u={ (3), (0) :}`

So if we substitute into the integral, we get:

`I = int_0^3 \ ( 2u^2 ) / ( u^2+9 ) \ du`
`\ \ = 2 \ int_0^3 \ ( u^2 ) / ( u^2+9 ) \ du`
`\ \ = 2 \ int_0^3 \ ( u^2+9-9 ) / ( u^2+9 ) \ du`
`\ \ = 2 \ int_0^3 \ 1 -9 / ( u^2+9 ) \ du`
`\ \ = 2 \ int_0^3 \ 1 -9 / ( u^2+3^2 ) \ du`

Which we can readily integrate:

`I = 2 [u - 9/3arctan(u/3)]_0^3`
`\ \ = 2 { (3-3arctan1) - (0-3arctan0)}`
`\ \ = 2 (3-(3pi)/4)`
`\ \ = 6 - (3pi)/2`