Question

求助，高数

Answer 1

(1)求定积分[0，3]∫dx/(x³+9)
解：此题书写很困难，写出来也看不清楚，我把方法写出来，最后的积分请自己作.。
1/（x³+9)=1/{[(x+9^(1/3)][x²-9^(1/3)x+9^(2/3)]}(分母按立方和分解因式)
={1/[3×9^(2/3)]}{1/[x+9^(1/3)]-[x-2×9^(1/3)]/[x²-9^(1/3)x+9^(2/3)]}(拆成两项，前面的
1/[3×9^(2/3)]是配出来的一个系数。)
={1/[3×9^(2/3)]}{1/[x+9^(1/3)]-[x-9^(1/3)]/[x²-9^(1/3)x+9^(2/3)]+[9^(1/3)]/[x²-9^(1/3)x+9^(2/3)]}
(再拆成三项).后面就是常规的积分。
(2)求不定积分∫(sin5x-cos3x)dx
解：原式=(1/5)∫sin5xd(5x)-(1/3)∫cos3xd(3x)=-(1/5)cos5x-sin3x+C

Answer 2

∫(sin5x-cos3x)dx
=∫sin5xdx-∫cos3xdx
=(-1/5)cos5x+(-1/3)sin3x+C

∫dx/(x^3+a^3)=∫dx/(x+a)(x^2-ax+a^2)=∫(1/3a^2)dx/(x+a)+∫[(-1/3a^2)x+(2/3a)]dx/(x^2-ax+a^2)
=(1/3a^2)ln|x+a| -(1/6a^2)∫d(x^2-ax+a^2)/(x^2-ax+a^2)+(1/3a)∫dx/(x^2-ax+a^2)
=(1/3a^2)ln|x+a|-(1/6a^2)ln|x^2-ax+a^2|+(2/(3√3a^2))∫d(2x/√3a-1/√3)/[(2x/√3-1/√3)^2+1]
=(1/3a^2)ln|x+a|-(1/6a^2)ln|x^2-ax+a^2|+[2/(3√3a^2)]arctan(2x/√3a-1/√3)
a=9^(1/3)
∫[0,3]dx/(x^3+9)=(1/3*9^2/3)ln(|3+9^(1/3)|/9^(1/3)|)
-(1/6*9^(2/3))ln(|9-3*9^(1/3)+9^(2/3)|/9^(2/3)| )
+(2/(3√3 *9^(2/3)))[arctan(6/√3*9^(1/3)-1/√3) -arctan)(-1/√3)]

1/(x^3+a^3)=A/(x+a)+(Bx+C)/(x^2-ax+a^2)
A(x^2-ax+a^2)+(Bx+C)(x+a)
=(A+B)x^2+(-Aa+aB+C)x+Aa^2+Ca
A+B=0 -A+aB+C=0 C=2aA
Aa^2+Ca=1 Aa^2+2a^2A=1
A=1/3a^2 B=-1/3a^2 C=2/3a