Set u=sqrtx>0u=√x>0 hence du=1/(2sqrtx)dx=>dx=2ududu=12√xdx⇒dx=2udu
Hence the integral becomes
int sqrtx/(sqrtx-3)dx=int 2*u^2/(u-3)du∫√x√x−3dx=∫2⋅u2u−3du
Analyze 2u^2/(u-3)2u2u−3 in partial fractions we get
int 2u^2/(u-3)du=int (2u+18/(u-3)+6)du=
u^2+18ln(u-3)+6u+c=x+18*ln(sqrtx-3)+6sqrtx+c∫2u2u−3du=∫(2u+18u−3+6)du=u2+18ln(u−3)+6u+c=x+18⋅ln(√x−3)+6√x+c
Finally
int sqrtx/(sqrtx-3)dx=x+18*ln(sqrtx-3)+6sqrtx+c∫√x√x−3dx=x+18⋅ln(√x−3)+6√x+c