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(2)
∫x/√(x^2+9) dx
=(1/2)∫d(x^2+9)/√(x^2+9)
=√(x^2+9) +C
(3)
∫(0->π/8) (sec2x)^2 dx
=(1/2) [ tan2x] |(0->π/8)
=1/2
(3)
∫ (2x-1)/(x^2-5x+6) dx
=∫ (2x-1)/[(x-2)(x-3)] dx
= ∫ [-3/(x-2) +5/(x-3)] dx
= -3ln|x-2| +5ln|x-3| + C
let
(2x-1)/[(x-2)(x-3)]≡ A/(x-2) +B/(x-3)
=>
2x-1≡ A(x-3) +B(x-2)
x=2 , =>A=-3
x=3, =>B=5
∫x/√(x^2+9) dx
=(1/2)∫d(x^2+9)/√(x^2+9)
=√(x^2+9) +C
(3)
∫(0->π/8) (sec2x)^2 dx
=(1/2) [ tan2x] |(0->π/8)
=1/2
(3)
∫ (2x-1)/(x^2-5x+6) dx
=∫ (2x-1)/[(x-2)(x-3)] dx
= ∫ [-3/(x-2) +5/(x-3)] dx
= -3ln|x-2| +5ln|x-3| + C
let
(2x-1)/[(x-2)(x-3)]≡ A/(x-2) +B/(x-3)
=>
2x-1≡ A(x-3) +B(x-2)
x=2 , =>A=-3
x=3, =>B=5
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