Question

高数定积分问题

高数定积分问题求此题证明

Answer 1

分段函数求积分
∫(-1→1) f(x)dx
=∫(0→1) ln(1+x)dx+∫(-1→0) xe^x² dx
先求∫ln(1+x)dx分部积分法：
∫ln(1 + x) dx
= x * ln(1 + x) - ∫x dln(1 + x)
= xln(1 + x) - ∫x / (1 + x) dx
= xln(1 + x) - ∫(1 + x - 1) / (1 + x) dx
= xln(1 + x) - ∫ dx + ∫ dx / (1 + x)
= xln(1 + x) - x + ln(x+1) + C |(0→1)
=ln2-1+ln2
=2ln2-1
再求∫xe^x² dx
令x²=t,dt=2xdx，xdx=dt/2
上式
=∫e^x²* xdx
=∫e^t *dt/2
=e^t/2+C
=e^x²/2+C |(-1→0)
=e^0-e^1/2
=1-√e
所以原式=2ln2-1+1-√e=2ln2-√e

Answer 2

∫1/(x²-a²)²dx=1/2∫1/[x(x²-a²)²]d(x²-a²)=-1/2∫1/xd1/(x²-a²)
=-1/[2x(x²-a²)]+1/2∫1/(x²-a²)d1/x
=-1/[2x(x²-a²)]-1/2∫1/[x²(x²-a²)]dx
=-1/[2x(x²-a²)]-1/(2a²)∫1/(x²-a²)-1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)∫1/(x-a)-1/(x+a)dx+1/(2a²)∫1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)ln|(x-a)/(x+a)|-1/(2a²x)+C