Question #8f777

1 Answer

Set u=sqrtx>0u=x>0 hence du=1/(2sqrtx)dx=>dx=2ududu=12xdxdx=2udu

Hence the integral becomes

int sqrtx/(sqrtx-3)dx=int 2*u^2/(u-3)duxx3dx=2u2u3du

Analyze 2u^2/(u-3)2u2u3 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+c2u2u3du=(2u+18u3+6)du=u2+18ln(u3)+6u+c=x+18ln(x3)+6x+c

Finally

int sqrtx/(sqrtx-3)dx=x+18*ln(sqrtx-3)+6sqrtx+cxx3dx=x+18ln(x3)+6x+c