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

Question

6583 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-24T22:32:24

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

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:

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

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