前置知识：定积分的计算（分部积分法）
习题1
计算 ∫ 1 3 arctan x d x \int_1^{\sqrt 3}\arctan xdx ∫13arctanxdx
解：
\qquad 原式 = x arctan x ∣ 1 3 − ∫ 1 3 x d ( arctan x ) = arctan 3 − arctan 1 − ∫ 1 3 x 1 + x 2 d x =x\arctan x\bigg\vert_1^{\sqrt 3}-\int_1^{\sqrt 3}xd(\arctan x)=\arctan \sqrt 3-\arctan 1-\int_1^{\sqrt 3}\dfrac{x}{1+x^2}dx =xarctanx
13−∫13xd(arctanx)=arctan3−arctan1−∫131+x2xdx
π 12 − 1 2 ln ( x 2 + 1 ) ∣ 1 3 = π 12 − 1 2 ln 2 \qquad\qquad \dfrac{\pi}{12}-\dfrac 12\ln(x^2+1)\bigg\vert_1^{\sqrt 3}=\dfrac{\pi}{12}-\dfrac 12\ln 2 12π−21ln(x2+1) 13=12π−21ln2
习题2
计算 ∫ 1 2 x ln x d x \int_1^2x\ln xdx ∫12xlnxdx
解：
\qquad 原式 = 1 2 ∫ 1 2 ln x d ( x 2 ) = 1 2 x 2 ln x ∣ 1 2 − 1 2 ∫ 1 2 x 2 d ( ln x ) =\dfrac 12\int_1^2\ln xd(x^2)=\dfrac 12x^2\ln x\bigg\vert_1^2-\dfrac 12\int_1^2x^2d(\ln x) =21∫12lnxd(x2)=21x2lnx
12−21∫12x2d(lnx)
= 2 ln 2 − 1 2 ∫ 1 2 x 2 ⋅ 1 x d x = 2 ln 2 − 1 2 ∫ 1 2 x d x \qquad\qquad =2\ln 2-\dfrac 12\int_1^2x^2\cdot \dfrac 1xdx=2\ln 2-\dfrac 12\int_1^2xdx =2ln2−21∫12x2⋅x1dx=2ln2−21∫12xdx
= 2 ln 2 − 1 4 x 2 ∣ 1 2 = 2 ln 2 − 3 4 \qquad\qquad =2\ln 2-\dfrac 14x^2\bigg\vert_1^2=2\ln 2-\dfrac 34 =2ln2−41x2 12=2ln2−43
习题3
设 f ′ ′ ( x ) f''(x) f′′(x)在 [ 0 , 1 ] [0,1] [0,1]上连续，且 f ( 0 ) = 1 , f ( 2 ) = 3 , f ′ ( 2 ) = 5 f(0)=1,f(2)=3,f'(2)=5 f(0)=1,f(2)=3,f′(2)=5，求 ∫ 0 1 x f ′ ′ ( 2 x ) d x \int_0^1xf''(2x)dx ∫01xf′′(2x)dx
解：
\qquad 原式 = 1 2 ∫ 0 1 x d [ f ′ ( 2 x ) ] = 1 2 x f ′ ( 2 x ) ∣ 0 1 − 1 2 ∫ 0 1 d [ f ′ ( 2 x ) ] =\dfrac 12\int_0^1xd[f'(2x)]=\dfrac 12xf'(2x)\bigg\vert_0^1-\dfrac 12\int_0^1d[f'(2x)] =21∫01xd[f′(2x)]=21xf′(2x)
01−21∫01d[f′(2x)]
= 1 2 f ′ ( 2 ) − 1 4 f ( 2 x ) ∣ 0 1 = 5 2 − 1 2 = 2 \qquad\qquad =\dfrac 12f'(2)-\dfrac 14f(2x)\bigg\vert_0^1=\dfrac 52-\dfrac 12=2 =21f′(2)−41f(2x) 01=25−21=2
总结
只要熟练地掌握分部积分法，这类题目一般都可以很轻松地解决。