How do you use partial fractions to find the integral #int (sec^2x)/(tanx(tanx+1))dx#?

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Answer 1 · 2017-01-30T20:36:16

#int sec^2x/(tanx(tanx+1))dx = ln abs (sinx/(sinx+cosx)) +C#

Substitute:

#t = tanx#

#dt = sec^2x dx#

so we have:

#int sec^2x/(tanx(tanx+1))dx = int (dt)/(t(t+1))#

Now we can solve the integral using partial fractions:

#1/(t(t+1)) = A/t+B/(t+1)#

#1/(t(t+1)) = (A(t+1)+ Bt)/(t(t+1))#

#1= (A+B)t +A#

#{(A = 1),(B=-1):}#

#int (dt)/(t(t+1)) = int (1/t-1/(t+1))dt#

#int (dt)/(t(t+1)) = int (dt)/t-int (dt)/(t+1)#

#int (dt)/(t(t+1)) = ln abs(t) -ln abs(t+1) +C =ln abs (t/(t+1)) +C#

and substituting back #x#:

#int sec^2x/(tanx(tanx+1))dx = ln abs (tanx/(tanx+1)) +C#

#int sec^2x/(tanx(tanx+1))dx = ln abs (sinx/(sinx+cosx)) +C#