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