Question

∫_(-1)^1▒〖√(1+x^2)〗dx 怎么算

Answer 1

被陪昌迹积函数是偶函数
原式= 2∫(0到迅和1) √(x²+1) dx
令x=tanψ，dx=sec²ψ dψ
当x=0，ψ=0 // 当x=1，ψ=π/4
原式= 2∫(0到π/4) (secψ)(sec²ψ) dψ = 2I
I = ∫(0到π/4) sec³ψ dψ
= ∫(0到π/4) secψ dtanψ
= (secψ*tanψ)[0到π/4] - ∫(0到π/4) tanψ d(secψ)
= √2 - ∫(0到π/4) tan²ψ*secψ dψ
= √2 - ∫(0到π/4) (sec²ψ-1)*secψ dψ
= √2 - I + ∫(0到π/4) secψ dψ
2I = √2 + ln|secψ+tanψ|[0到π/4]
I = √2/2 + (1/2)ln(√2+1) - (1/2)ln(1+0) = √2/2 + (1/2)ln(1+√2)
原式= 2I = √2 + ln(1+√2)

∴∫(-1到1) √(x²芦并+1) dx = √2 + ln(1+√2)