Question

求不定积分(1) (6-x)/((9-x^2))dx

Answer 1

问：求不定积分(1) (6-x)/((9-x^2))dx

问：分母加根号

答：∫(6-x)/√(9-x^2)dx=∫6/√(9-x²)-x/√(9-x²)dx=-6arcos(x/3)-1/2∫1/√(9-x²)d(x²)=-6arcos(x/3)+(9-x²)½

答：∫(6-x)/√(9-x^2)dx=∫6/√(9-x²)-x/√(9-x²)dx=-6arcos(x/3)-1/2∫1/√(9-x²)d(x²)=-6arcos(x/3)+(9-x²)^1/2+C

答：解析:∫6/√(9-x²)dx=-arcos(x/3)这里可以用换元法令x=3cost原式=∫1/[√(9-9costcost)]d(3cost)=∫sint^(-1)(-3sint)/3dt=-∫dt=-t+C=-arccos(x/3)+C

答：解析:∫6/√(9-x²)dx=-6arcos(x/3)这里可以用换元法令x=3cost原式=∫6/[√(9-9costcost)]d(3cost)=∫6sint^(-1)(-3sint)/3dt=-∫6dt=-6t+C=-6arccos(x/3)+C

答：不好意思，上面少打一个6，过程没问题，主要是换元法

问：这个的不定积分老师

答：∫(x+1)/(4+x²)dx=∫x/(4+x²)dx+∫1/(4+x²)dx=1/2∫1/(4+x²)d(x²)+1/2arctan(x/2)+C=1/2ln(4+x²)+1/2arctan(x/2)+C