Question

How to integrate ?

Answer 1

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))])`