Question

Evaluate the integral `int (xlnx)/sqrt(x^2-1)`?

Answer 1

`int (xlnx)/sqrt(x^2-1) = lnxsqrt(x^2-1) - sqrt(x^2-1) + arctansqrt(x^2-1) + C`

Explanation:

We seek:

`I = int (xlnx)/sqrt(x^2-1)`

Let `{ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=x/sqrt(x^2-1), => v,=sqrt(x^2-1) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (lnx)(x/sqrt(x^2-1)) \ dx = (lnx)(sqrt(x^2-1)) - int \ (sqrt(x^2-1))(1/x) \ dx`

`:. I = lnxsqrt(x^2-1) - int \ sqrt(x^2-1)/x \ dx` ..... [A]

For the next integral, `int \ sqrt(x^2-1)/x \ dx`, we can perform a substitution.

Let `u=sqrt(x^2-1) => (du)/dx=x/sqrt(x^2-1)`

Then substituting intio the last integral, we get:

`int \ sqrt(x^2-1)/x \ dx = int \ u^2/(u^2+1) \ du`
`" " = int \ (u^2+1-1)/(u^2+1) \ du`
`" " = int \ 1 - 1/(u^2+1) \ du`
`" " = u - arctanu`

Restoring the substitution and using the result with [A] we get:

`I = lnxsqrt(x^2-1) - {sqrt(x^2-1) - arctansqrt(x^2-1) } + C`
`\ \ = lnxsqrt(x^2-1) - sqrt(x^2-1) + arctansqrt(x^2-1) + C`