Integration by parts involving exponentials and logarithms

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SUMMARY

The discussion focuses on the integration of the function (1/x^2)(lnx) using integration by parts, specifically within the limits of e and 1. The user correctly identifies u as x^-2 and calculates du/dx as -2x^-3, while dv/dx is determined to be lnx, leading to v = x(lnx - 1). The integration process involves substituting these values into the integration by parts formula, ultimately leading to a successful resolution of the integral.

PREREQUISITES
  • Understanding of integration by parts formula: ∫udv = uv - ∫vdu
  • Knowledge of logarithmic functions and their properties
  • Familiarity with differentiation techniques, particularly for power functions
  • Basic skills in evaluating definite integrals
NEXT STEPS
  • Practice additional integration by parts problems involving logarithmic and exponential functions
  • Explore the properties of definite integrals and their applications
  • Learn about alternative integration techniques such as substitution and partial fractions
  • Study the behavior of logarithmic functions in calculus for deeper insights
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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of integration by parts involving logarithmic functions.

xllx
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Homework Statement



Using integration by parts, integrate:
(1/x^2)(lnx) dx with the limits e and 1


Homework Equations



[uv]to the limits a b - the integral of (v)(du/dx) dx

(sorry, don't know how to write out equations properly on a computer)

The Attempt at a Solution



I've got u=x^-2 so du/dx= -2x^-3
dv/dx=lnx so v= x(lnx-1)

So putting this into the equation above:
[x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1))

Is this right so far?
If so how do I integrate the last part? Do I do it sepeartley or by parts again?

Many Thanks, any help at all would be greatly appreciated.
 
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Hi xllx! :smile:

(have an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
xllx said:
Using integration by parts, integrate:
(1/x^2)(lnx) dx with the limits e and 1

I've got u=x^-2 so du/dx= -2x^-3
dv/dx=lnx so v= x(lnx-1)

So putting this into the equation above:
[x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1))

eugh!

go the other way …

integrate x-2 ! :smile:
 
Thankyou. Redid it and came out with a reasonal answer.
 

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