Indefinite Integral Calculus II

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SUMMARY

The forum discussion centers on solving the indefinite integral of the function \(\int \frac{ \tan x \sec^2 x }{ \tan^2 x + 6 \tan x + 8 } dx\). The user successfully substituted \(u = \tan x\) and \(du = \sec^2 x \, dx\), transforming the integral into \(\int \frac{ u }{ (u+4)(u+2) } du\). However, the user encountered difficulties while attempting to apply partial fraction decomposition, specifically with the expression \(\frac{2}{u+4} + \frac{-1}{u+2}\). The discussion seeks assistance in resolving this issue.

PREREQUISITES
  • Understanding of integral calculus, specifically indefinite integrals.
  • Familiarity with trigonometric identities, particularly tangent and secant functions.
  • Knowledge of substitution methods in integration.
  • Experience with partial fraction decomposition techniques.
NEXT STEPS
  • Review the method of substitution in integral calculus.
  • Study partial fraction decomposition in detail, focusing on its application in integrals.
  • Practice solving integrals involving trigonometric functions and their derivatives.
  • Explore advanced techniques in integration, such as integration by parts and trigonometric substitutions.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators seeking to enhance their teaching methods in trigonometric integrals.

Wm_Davies
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Homework Statement



\int \frac{ \tan x \sec^2 x }{ \tan^2 x + 6 \tan x + 8 } dx

Homework Equations





The Attempt at a Solution



\int \frac{ \tan x \sec^2 x }{ \tan^2 x + 6 \tan x + 8 } dx
Okay I let...

u=tanx
du=sec^2x


Then I got

\int \frac{ u }{ (u+4)(u+2) } du

Then I split this into partial fractions which is not working... for some reason. Can somebody please help me?
 
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\frac {2}{u+4} + \frac{-1}{u+2}
 

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