Integration by Substitution using Partial Fractions Decomposition

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SUMMARY

The discussion focuses on integrating the function \(\int\frac{dz}{1+e^z}\) using substitution and partial fractions decomposition. The user sets \(u = 1 + e^z\), leading to the transformation of the integral into \(\int\frac{1}{u} \frac{du}{e^{z}}\). The next step involves rewriting \(\frac{1}{u(u-1)}\) as a sum of partial fractions, specifically \(A/u + B/(u-1)\), and solving for constants \(A\) and \(B\) to facilitate integration.

PREREQUISITES
  • Understanding of integration techniques, specifically substitution.
  • Familiarity with exponential functions and their properties.
  • Knowledge of partial fractions decomposition.
  • Ability to manipulate algebraic expressions and solve for constants.
NEXT STEPS
  • Study the method of partial fractions decomposition in detail.
  • Learn about integration techniques involving exponential functions.
  • Practice solving integrals using substitution with various functions.
  • Explore advanced integration techniques, such as integration by parts.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of substitution and partial fractions decomposition in action.

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Homework Statement



Integrate [tex]\int[/tex][tex]\frac{dz}{1+e^z}[/tex] by substitution

Homework Equations





The Attempt at a Solution



I chose u=(1+[tex]e^{z}[/tex]) so du/dz=[tex]e^{z}[/tex] and dz=du/[tex]e^{z}[/tex].

Therefore, [tex]\int[/tex][tex]\frac{1}{u}[/tex] [tex]\frac{du}{e^{z}}[/tex]

I plug z=ln(u-1) in for z, so [tex]\int[/tex][tex]\frac{1}{u}[/tex] [tex]\frac{du}{u-1}[/tex]

From here though I don't know how to integrate. Can anyone help me with the next step?
 
Physics news on Phys.org
Rewrite 1/(u(u -1)) as a sum: A/u + B/(u - 1). Solve for A and B so that the two expressions are identically equal. This is called partial fractions decomposition.
 

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