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Question #58927. Asked by **blurrystar1**.

Last updated **Oct 23 2017**.

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19 year member

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The definite integral of lnx is xlnx - x. (lnx is "log of x to the base e")

Aug 21 2005, 6:11 AM

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Answer has

That's correct. To work it out, you have to use integration by parts: The integral of u*dv is equal to u*v - (the integral of v*du). This works whenever one part can be diffentiated but not integrated easily, just like lnx. In this case, lnx=u and dx=dv. Then you just plug those values into the formula to solve.

Aug 21 2005, 8:55 AM

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Answer has

if f(x) = xlnx

f'(x) = 1 + lnx (using product rule)

hence (assume antidif is the antidif sign, lil squiggly line wateva)

antidif (d xlnx/dx)dx = antidif(1)dx + antidif(lnx)dx

xlnx - antidif(1)dx = antidif(lnx)dx

xlnx - x + c = antidif(lnx)dx

f'(x) = 1 + lnx (using product rule)

hence (assume antidif is the antidif sign, lil squiggly line wateva)

antidif (d xlnx/dx)dx = antidif(1)dx + antidif(lnx)dx

xlnx - antidif(1)dx = antidif(lnx)dx

xlnx - x + c = antidif(lnx)dx

Jun 13 2008, 9:57 PM