∫x+3/x^2-x-2dx=?
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∫x+3/x^2-x-2dx=?∫x+3/x^2-x-2dx=?∫x+3/x^2-x-2dx=?(x+3)/(x²-x-2)=(x+3)/(x+1)(x-2)令A/(x+1)+B/(x-2)
∫x+3/x^2-x-2dx=?
∫x+3/x^2-x-2dx=?
∫x+3/x^2-x-2dx=?
(x+3)/(x²-x-2)
=(x+3)/(x+1)(x-2)
令A/(x+1)+B/(x-2)=(x+3)/(x+1)(x-2)
即(A(x-2)+B(x+1))/(x+1)(x-2)=(x+3)/(x+1)(x-2)
∴A+B=1 B-2A=3
解得A=-2/3 B=5/3
∴∫(x+3)/(x²-x-2)dx
=-2/3*∫1/(x+1)dx+5/3*∫1/(x-2)dx
=-2/3ln|x+1|+5/3ln|x-2|+C
=(5ln|x-2|-2ln|x+1|)/3+C
∫(x+3)/(x^2-x-2)dx
=∫(x+3)/(x-2)(x+1)dx
=∫[(5/3)/(x-2)+(-2/3)/(x+1)]dx
=(5/3)∫1/(x-2)dx-(2/3)∫1/(x+1)dx
=(5/3)∫1/(x-2)d(x-2)-(2/3)∫1/(x+1)d(x+1)
=(5/3)ln(x-2)-(2/3)ln(x+1)+C
有不懂欢迎追问