Question

Question #8f777

Answer 1

Set `u=sqrtx>0` hence `du=1/(2sqrtx)dx=>dx=2udu`

Hence the integral becomes

`int sqrtx/(sqrtx-3)dx=int 2*u^2/(u-3)du`

Analyze `2u^2/(u-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#

Finally

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