What is the improper integrals sqrtx ln 5x dx from 1 to e ?

1 Answer
May 11, 2018

=2/3e^(3/2)ln(5e)-4/9e^(3/2)-2/3ln(5)+4/9=23e32ln(5e)49e3223ln(5)+49

Explanation:

In preparation for evaluating the definite integral, we should first find the antiderivative intsqrtxln(5x)dxxln(5x)dx, which can be solved using Integration by Parts:

u=ln(5x)u=ln(5x)
du=x^-1dxdu=x1dx
dv=sqrtxdxdv=xdx
v=2/3x^(3/2)v=23x32

uv-intvdu=2/3x^(3/2)ln(5x)-2/3intx^(3/2)x^(-1)dxuvvdu=23x32ln(5x)23x32x1dx

#=2/3x^(3/2)ln(5x)-2/3intsqrtxdx#

=2/3x^(3/2)ln(5x)-4/9x^(3/2)=23x32ln(5x)49x32 (leaving out the constant as we're going to use this to evaluate a definite integral)

Now, we may evaluate the improper definite integral:

int_1^esqrtxln(5x)dxe1xln(5x)dx

This is not an improper integral; the integrand sqrt(x)ln(5x)xln(5x) is continuous on the interval of integration [1, e][1,e].

Thus,

int_1^esqrtxln(5x)dx=[2/3x^(3/2)ln(5x)-4/9x^(3/2)]|_1^ee1xln(5x)dx=[23x32ln(5x)49x32]e1

=2/3e^(3/2)ln(5e)-4/9e^(3/2)-2/3ln(5)+4/9=23e32ln(5e)49e3223ln(5)+49