`int (x^3-x+3)/(x^2+x-2) dx` Use partial fractions to find the indefinite integral

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

The given integrand is a improper rational function, as the degree of the numerator is more than the degree of the denominator. To apply the method of partial fractions,first we have to do a division with remainder.

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

Since the polynomials do not completely divide, we have to continue...

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

The given integrand is a improper rational function, as the degree of the numerator is more than the degree of the denominator. To apply the method of partial fractions,first we have to do a division with remainder.

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

Since the polynomials do not completely divide, we have to continue partial fractions on the remainder.

We need to factor the denominator,

`(2x+1)/(x^2+x-2)=(2x+1)/(x^2-x+2x-2)`

`=(2x+1)/(x(x-1)+2(x-1))`

`=(2x+1)/((x-1)(x+2))`

Now let's create the partial fraction template,

`(2x+1)/((x-1)(x+2))=A/(x-1)+B/(x+2)`

Multiply the above equation by denominator,

`=>2x+1=A(x+2)+B(x-1)`

`2x+1=Ax+2A+Bx-B`

`2x+1=(A+B)x+2A-B`

Equating the coefficients of the like terms,

`A+B=2`    ------------------------(1)

`2A-B=1`  -------------------------(2)``

Now we have to solve the above two linear equations to get A and B,

Add the equations 1 and 2,

`A+2A=2+1`

`3A=3`

`A=1`

Plug in the value of A in equation 1,

`1+B=2`

`B=2-1=1`

Now plug in the values of A and B in the partial fraction template,

`(2x+1)/((x-1)(x+2))=1/(x-1)+1/(x+2)`

Now we can evaluate the integral as,

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

Apply the sum rule,

`=intxdx-int1dx+int1/(x-1)dx+int1/(x+2)dx`

For the first and second integral apply the power rule and for the third and fourth integral use the common integral:`int1/xdx=ln|x|`

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

Add a constant C to the solution,

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

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