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#