How to integrate ?

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Answer 1 · 2015-12-16T15:34:40

Factorize the denominator

#int1/(x(x^2+3))dx#

do partial fraction

#1/(x(x^2+3))=A/x+(Bx+C)/(x^2+3)#

Multiply both side by #x(x^2+3)#

#1=A(x^2+3)+(Bx+C)x#

#1=Ax^2+3A+Bx^2+Cx#

#1=x^2(A+B)+Cx+3A#

#A+B = 0#
#C = 0#
#3A = 1#

#A = 1/3#
#B = -1/3#
#C = 0#

#1/(x(x^2+3)) = 1/(3x)-x/(3(x^2+3)#

So

#int1/(x(x^2+3))dx = 1/3(int1/(x)dx-intx/(x^2+3)dx)#

#1/3([ln(x)]-intx/(x^2+3)dx)#

#=intx/(x^2+3)dx = 1/2int(2x)/(x^2+3)dx#

let's #u = x^2+3#
so #du = 2xdx#

#1/2int1/udu = [ln(sqrt(u))]#

substitute back

#int1/(x(x^2+3))dx = 1/3([ln(x)]-[ln(sqrt(x^2+3))])#