Integration

Solution

For this question, we need to use substitution to integrate this question,

∫[1/(x^2 + 4)] dx
Put x = 2tanθ,
Then dx = 2(secθ)^2 dθ
Also, θ = tan-¹(x/2)

The given integral becomes
= ∫ 2(secθ)^2/ [(2tanθ)^2 + 4] dθ
= ∫ 2(secθ)^2/ [4(tanθ)^2 + 4] dθ
= ∫ 2(secθ)^2/ [4(secθ)^2] dθ
= ∫½ dθ
= ½∫dθ
= ½ (θ) + C
= tan-¹(x/2) + C, C is a constant

∫ {1/[(x^2)+a^2]}=ln|x+√(x^2+a^2)|+C

∫ {1/[(x^2)+4]}
=ln|x+√(x^2+4)|+C

係咪1呀?

2006-11-02 11:59:27 補充：
你in一個-1次方ge野,in 完咪0次方囉!