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求助根号下(X-5/X+2)的不定积分怎么做啊?
题目详情
【求助】根号下(X-5/X+2)的不定积分怎么做啊?
▼优质解答
答案和解析
令(x-5)/(x+2)=u²
则x=(2u²+5)/(1-u²)
dx=[(4u(1-u²)+2u(2u²+5)]/(1-u²)²du=14u/(1-u²)²du
原式=∫14u²/(1-u²)²du
=7/2∫[1/(1-u)²+1/(1+u)²-1/(1-u)-1/(1+u)]du
=7/2[1/(1-u)-1/(1+u)+ln|1-u|-ln|1+u|]+C
代入u即可.
则x=(2u²+5)/(1-u²)
dx=[(4u(1-u²)+2u(2u²+5)]/(1-u²)²du=14u/(1-u²)²du
原式=∫14u²/(1-u²)²du
=7/2∫[1/(1-u)²+1/(1+u)²-1/(1-u)-1/(1+u)]du
=7/2[1/(1-u)-1/(1+u)+ln|1-u|-ln|1+u|]+C
代入u即可.
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