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`int (x^2 - x + 6)/(x^3 + 3x) dx` Evaluate the integral

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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