Integrate `int(x^2-x+6)/(x^3+3x)`

Rewrite the rational function using partial fractions.

`(x^2-x+6)/(x^3+3x)=(A/x)+(Bx+C)/(x^2+3)`

`x^2-x+6=A(x^2+3)+(Bx+C)x`

`x^2-x+6=Ax^2+3A+Bx^2+Cx`

`x^2-x+6=(A+B)x^2+Cx+3A`

Equate coefficients and solve for A, B, and C.

`3A=6`

`A=2`

`C=-1`

`A+B=1`

`2+B=1`

`B=-1`

`int(x^2-x+6)/(x^3+3x)dx=int(2/x)dx+int[(x-1)/(x^2+3)]dx`

`=int(2/x)dx+int[x/(x^2+3)]dx-int[1/(x^2+3)]dx`

The first integral matches the form `int (du)/u=ln|u|+C`

`int(2/x)dx=2int(1/x)dx=2ln|x|+C`

Integrate the second integral using u-substitution.

Let `u=x^2+3`

`(du)/(dx)=2x``

`dx=(du)/(2x)`

` ` `-intx/(x^2+3)dx`

`=-int(x/u)*(du)/(2x)`

`=-1/2ln|u|`

`=-1/2ln|x^2+3|`

The third integral matches the form

`int(dx)/(x^2+a^2)=(1/a)tan^-1(x/a)+C`

`-int1/(x^2+3)dx`

`=-1/sqrt3tan^-1(x/sqrt3)+C`

**The final answer is:**

`2ln|x|-1/2ln|x^2+3|-1/sqrt3tan^-1(x/sqrt3)+C`

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