定積分の問題

質問

∫[1,√3] arctanx/x^2 dx
はどのように計算しますか？

回答No.2

bran111 · ベストアンサー

I=∫[1,√3] arctanx/x^2 dx

t=arctanx、x=tant

と置く。

dx/dt=1/cos^2t,

積分範囲　：[1,√3]　⇒　[π/4,π/3]

I=∫[π/4,π/3] [t/tan^2t]dt/cos^2t=∫[π/4,π/3] [t/sin^2]dt=∫[π/4,π/3]f(t)g'(t)dt

とおいて部分積分を行う。

f(t)=t, g'(t)=1/sin^2t　⇒　f'(t)=1,　g(t)：以下に求める。

g(t)=∫(1/sin^2)dt

天下りだが

d(cost/sint)dt=-1/sin^2t (実際に微分すればすぐ解る)

よって

g(t)=∫(1/sin^2)dt＝-cost/sint

I=∫[π/4,π/3]f(t)g'(t)dt=[f(t)g(t)][π/4,π/3]-∫[π/4,π/3]f'(t)g(t)dt
=[-tcost/sint][π/4,π/3]-∫[π/4,π/3](-cost/sint)dt
=-[π/3(1/2)/(√3/2)-π/4(1/√2)/(1/√2)]+∫[π/4,π/3](cost/sint)dt
=π(9-4√3)/36+J

J=∫[π/4,π/3](cost/sint)dt=∫[π/4,π/3](d(sint)/dt)/sint)dt=∫[π/4,π/3](d(sint)/sint)
=[log(sint)][π/4,π/3]=log[(√3/2)/(1/√2)]=log(√3/√2)=(1/2)log(3/2)

I=π(9-4√3)/36+(1/2)log(3/2)

回答No.1

info222_

I=∫[1,√3] arctan(x)/x^2 dx
部分積分して
=[-(1/x)arctan(x)][1,√3]+∫[1,√3] 1/(x(1+x^2)) dx
=π/4-(1/√3)π/3+∫[1,√3] ((1/x)-x/(1+x^2)) dx
=π(9-4√3)/36+[log|(x)-(1/2)log(1+x^2)][1,√3]
=π(9-4√3)/36+[log|(√3)-(1/2)log(4)+(1/2)log(2)]
=π(9-4√3)/36+(1/2)log(3/2) ...(答)