Question #8f777

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Answer 1 · 2016-05-28T16:19:19

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

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