- 求解以下不定积分题1)∫ln(2x
- 1)∫ln(2x - 3) dx
2)∫2 sin(2x -3) dx
3)∫1/(3-2x)dx
4)∫(3-2x)3次方dx
5)∫tan(2x-3)dx
6)∫1/3(1+x方)dx
- 1、分部积分:
∫ln(2x-3) dx
=xln(2x-3)-∫2x/(2x-3) dx
=xln(2x-3)-∫[1+3/(2x-3)]dx
=xln(2x-3)-x-3/2×∫1/(2x-3)d(2x-3)
=xln(2x-3)-x-3/2×ln(2x-3)+
2、∫2sin(2x-3)dx=∫sin(2x-3)d(2x-3)=-cos(2x-3)+C
3、∫1/(3-2x)dx=-1/2×∫1/(3-2x)d(3-2x)=-1/2×ln|3-2x|+C
4、∫(3-2x)^3 dx=-1/2×∫(3-2x)^3 d(3-2x)=-1/2×1/4×(3-2x)^4+C=-1/8×(3-2x)^4+C
5、因为∫tanx dx=-ln|cosx|+C,所以,
∫tan(2x-3)dx=1/2×∫tan(2x-3)d(2x-3)=-1/2×ln|cos(2x-3)|+C
6、是1/[3(1+x^2)],还是1/3×(1+x^2) ??
∫1/[3(1+x^2)]dx=1/3×∫1/(1+x^2) dx=1/3×arctanx+C;
∫1/3×(1+x^2)dx=1/3×(x+x^3/3)+C=1/3×x+1/9×x^3+C。