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What is antiderivative (1-lnx)/x^3?

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xav1394 | Student, Grade 11 | (Level 2) eNoter

Posted December 28, 2011 at 2:20 AM via web

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What is antiderivative (1-lnx)/x^3?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 28, 2011 at 2:47 AM (Answer #1)

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You need to start from the given derivative of the function and go to the original function. The operation you need to use is integration.

`int ((1-lnx)dx)/x^3 = int dx/x^3 - int (ln x dx)/x^3`

You need to solve the integrals separately such that:

`int dx/x^3 = int x^(-3)dx = (x^(-3+1))/(-3+1) + c`

`int dx/x^3 = -1/(2x^2) + c`

You need to use integration by parts to solve int (ln x dx)/x^3.

`u =ln x=gt du = dx/x`

`` `dv = dx/x^3 =gt v =-1/(2x^2)`

Use the formula of integration by parts:

`int udv = uv - int vdu`

`` `int (ln x dx)/x^3 = -ln x/(2x^2) + int dx/(2x^3)`

`int (ln x dx)/x^3 = -ln x/(2x^2) + (1/2)*(-1/(2x^2)) + c`

`int (ln x dx)/x^3 = (-1/(2x^2))*(ln x + 1/2) + c`

Evaluating the antiderivative of the given function yields:`int (ln x dx)/x^3 = (-1/(2x^2))*(ln x + 1/2) + c.`

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