求解不定积分：∫(x^5)/(Cx+D)dx, C,D 为常数。
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解决时间 2021-04-23 06:53
- 提问者网友：挣扎重来
- 2021-04-22 14:04
求解不定积分：∫(x^5)/(Cx+D)dx, C,D 为常数。
最佳答案
- 二级知识专家网友：眠于流年
- 2021-04-22 15:34
求解不定积分：∫(x⁵)/(Cx+D)dx, C,D 为常数。
解：当c≠0时：
∫{[x⁵/(cx+d)]dx=∫[(1/c)x⁴-(d/c²)x³+(d²/c³)x²-(d³/c⁴)x+(d⁴/c⁵)]-d⁵/[c⁴(cx+d)]}dx
=(1/5c)x⁵-(d/4c²)x⁴+(d²/3c³)x³-(d³/2c⁴)x²+(d⁴/c⁵)x-(d⁵/c⁵)ln∣cx+d∣+C
当c=0时∫(x⁵/d)dx=x⁶/(6d)+C.
解：当c≠0时：
∫{[x⁵/(cx+d)]dx=∫[(1/c)x⁴-(d/c²)x³+(d²/c³)x²-(d³/c⁴)x+(d⁴/c⁵)]-d⁵/[c⁴(cx+d)]}dx
=(1/5c)x⁵-(d/4c²)x⁴+(d²/3c³)x³-(d³/2c⁴)x²+(d⁴/c⁵)x-(d⁵/c⁵)ln∣cx+d∣+C
当c=0时∫(x⁵/d)dx=x⁶/(6d)+C.
全部回答
- 1楼网友：旧事诱惑
- 2021-04-22 17:27
因为f(x)=ax7+bx5+cx3+dx+5，
所以
f(-7)=a(-7)^7+b(-7)^5+c(-7)^3+d(-7)+5=-7
a(-7)^7+b(-7)^5+c(-7)^3+d(-7)=-7-5=-12
即
a(7)^7+b(7)^5+c(7)^3+d(7)=12
从而
f(7)=a(7)^7+b(7)^5+c(7)^3+d(7)+5=12+5=17.
- 2楼网友：堕落奶泡
- 2021-04-22 16:10
如果C=0，∫(x^5)/(Cx+D)dx=x^6/6D+A,下设C不为0
∫(x^5)/(Cx+D)dx
=(1/C)∫(x^5+(D/C)^5-(D/C)^5)/(x+D/C)dx
=(1/C)∫(x^4-x^3(D/C)+x^2(D/C)^2-x(D/C)^3+(D/C)^4)dx-∫(D/C)^5)/(x+D/C)dx
=(1/C)(x^5/5-x^4/4(D/C)+x^3/3(D/C)^2-x^2/2(D/C)^3+x(D/C)^4)-(D/C)^5)ln|x+D/C|+A
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