`int (4x^2+2x-1)/(x^3+x^2)dx`

To solve using partial fraction method, the denominator of the integrand should be factored.

`(4x^2+2x-1)/(x^3+x^2)=(4x^2+2x-1)/(x^2(x+1))`

Take note that if the factor in the denominator is linear and non-repeating, each factor in the denominator has a partial fraction form of `A/(ax+b)` .

And if the factor is linear and repeating, its partial fraction decomposition has a form `A_1/(ax+b) + A_2/(ax+b)^2+... +A_n/(ax+b)^n` .

So, expressing the integrand as sum of fractions, it becomes:

`(4x^2+2x-1)/(x^2(x+1))=A/x+B/x^2+C/(x+1)`

To determine the values of A, B and C, multiply both sides by the LCD of the fractions present.

`x^2(x+1)*(4x^2+2x-1)/(x^2(x+1))=(A/x+B/x^2+C/(x+1))*x^2(x+1)`

`4x^2+2x-1=Ax(x+1)+B(x+1)+Cx^2`

Then, assign values to x in which either x, x^2 or x+1 will become zero.

So plug-in x=0 to get the value of B.

`4(0)^2+2(0)-1=A(0)(0+1)+B(0+1)+C(0)^2`

`-1=A(0)+ B(1)+C(0)`

`-1=B`

Also, plug-in x=-1 to get the value of C.

`4(-1)^2+2(-1)-1=A(-1)(-1+1)+B(-1+1)+C(-1)^2`

`1=A(0)+B(0)+C(1)`

`1=C`

To get the value of A, plug-in the values of B and C. Also, assign any value to x. Let it be x=1.

`4(1)^2+2(1)-1=A(1)(1+1)+ (-1)(1+1)+1(1)^2`

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

`5=2A-1`

`6=2A`

`3=A`

So, the partial fraction decomposition of the integrand is:

`int(4x^2+2x-1)/(x^3+x^2)dx`

`= int (4x^2+2x-1)/(x^2(x+1))dx`

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

Then, express it as three integrals.

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

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

For the first and third integral, apply the formula `int 1/u du = ln|u|+C` .

And for the second integral, apply the formula `int u^n du = u^(n+1)/(n+1)+C` .

`= 3ln|x| + x^(-1) + ln|x+1| +C`

`=3ln|x| +1/x + ln|x+1| +C`

**Therefore, `int(4x^2+2x-1)/(x^3+x^2)=3ln|x| +1/x + ln|x+1| +C` .**

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