Evaluate: ∫sec^2 x/(tan^2 x + 4tanx)dx ​

Question

Evaluate:

\(\int\frac{see^2x}{(tan^3x+4tanx)}dx\) 

∫sec²x/(tan²x + 4tanx)dx

Answer 1

Let I = \(\int\frac{see^2x}{(tan^3x+4tanx)}dx\)

Putting t = tan x 

dt = sec² xdx

A(t² + 4) + (Bt + C)t = 1 

Putting t = 0, 

A(0 + 4) × B(0) =1 

A = 1/4

By equating the coefficients of t² and constant here, 

A + B = 0

1/4 + B = 0

B = - 1/4, C = 0

Now From equation (1) we get,