Solutions for these integration problems?

Question

∫[log(cosx)/cos²x].dx = ?

∫[(4x²-3x+2)/(4x²-4x+3)].dx = ?

Answer 1

1) ∫[log(cosx)/cos²x].dx = ∫[log(cosx) sec² x]dx

By parts, put u = log(cosx) , dv = sec² x, so that v = tan x

∫[log(cosx)/cos²x].dx = ∫ u dv = uv - ∫ v du = log(cosx) tan x - ∫ tan x (-sin x/ cos x) dx = log(cosx) tan x + ∫ tan² x dx = log(cosx) tan x + ∫ (sec² x -1) dx = log(cosx) tan x + tan x - x + C

2)∫[(4x²-3x+2)/(4x²-4x+3)].dx

(4x²-3x+2)/(4x²-4x+3) = (4x²-4x+3 + x - 1)/(4x²-4x+3) = 1 - (x -1)/(4x²-4x+3) = 1 - (1/8) (8x - 4 - 4)/(4x²-4x+3) = 1 -(1/8) (8x - 4)/(4x²-4x+3) + (1/2)/(4x²-4x+3)

The integrals of the 1st 2 parcels are x and (-1/8) ln((4x²-4x+3)
To integrate 1/(4x²-4x+3), observe that 4x²-4x+3 = (1/4) (x² - x + 3/4) = (1/4) [(x -1/2)² + 1/2] =(1/4)[(x -1/2)² + (sqrt(1/2))² ]
So, ∫(1/2)/(4x²-4x+3) = (1/2) * (1/4) * 1/sqrt(1/2) arctan(x -1/2)/sqrt(1/2)).

Now, just add the 3 parcels and do some algebra
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Answer 2

1.
∫[log(cosx)/cos²x].dx =∫log(secx).dx=(cosx)X(sinx)=(1/2)(sin2x)

2.don't know sorry

Answer 3

http://integrals.wolfram.com/index.jsp
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