Partial Fraction Decomposition for Integrating Rational Functions

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JamesGoh
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Im trying to find the integral of ( sec(t)^2 ) / ( (tan(t)^3) + (tan(t)^2) ). I've managed to get the
integral into the form

1 / (u^3 + u^2) where u = tan(t), however I am having difficulty proceeeding from there.

Could someone take a look at the working out I have attached and let me know what I am not doing right? (the correct answer is written in red pen on 2nd page)
 
Attachments
  • Handwritten calculus working on lined paper: integral of sec^2 t over (tan^3 t + tan^2 t) dt, rewritten with u = tan t, then partial fraction setup 1/(u^2(u+1)) = A/u^2 + B/(u+1) + C/u.
    Tute2Q5Hp1.jpg
    27.1 KB · Views: 493
  • Handwritten partial fraction working on lined paper: integral of 1/(u^2(u+1)) du split as -1/(u(u+1)) plus integral of 1/(u(u+1)), with a substitution v = u+1 and A/V + B/(V-1)^2 setup, and a red-pen result beside it.
    Tute2Q5Hp2.jpg
    29.1 KB · Views: 497
In order to evaluate ##\int\frac{1}{u^2(u+1)}\ du##, you want to use partial fraction decomposition, and that alone. You do not need to do any integration by parts. You have a correct general form $$\frac{1}{u^2(u+1)}=\frac{A}{u^2}+\frac{B}{u+1}+\frac{C}{u}$$ for the PFD, but then it looks like it all goes south after that, and you just gave up on that idea. Stick with that Idea.