`int (x^2-1)/(x^3+x) dx` Use partial fractions to find the indefinite integral

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

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

Now let's create partial fraction template,

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

Multiply equation by the denominator,

`(x^2-1)=A(x^2+1)+(Bx+C)x`

`(x^2-1)=Ax^2+A+Bx^2+Cx`

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

Comparing the coefficients of the like terms,

`A+B=1`  ----------------(1)

`C=0`

`A=-1`

Plug the value of A in equation 1,

`-1+B=1`

`B=2`

Plug in the values of A,B and C in the partial fraction template,

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

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

Apply the sum rule,

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

Take the constant out,

`=-1int1/xdx+2intx/(x^2+1)dx`

Now evaluate both the integrals separately,

`int1/xdx=ln|x|`

Now let's evaluate second integral,

...

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

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

Now let's create partial fraction template,

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

Multiply equation by the denominator,

`(x^2-1)=A(x^2+1)+(Bx+C)x`

`(x^2-1)=Ax^2+A+Bx^2+Cx`

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

Comparing the coefficients of the like terms,

`A+B=1`  ----------------(1)

`C=0`

`A=-1`

Plug the value of A in equation 1,

`-1+B=1`

`B=2`

Plug in the values of A,B and C in the partial fraction template,

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

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

Apply the sum rule,

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

Take the constant out,

`=-1int1/xdx+2intx/(x^2+1)dx`

Now evaluate both the integrals separately,

`int1/xdx=ln|x|`

Now let's evaluate second integral,

`intx/(x^2+1)dx`

Apply integral substitution: `u=x^2+1`

`du=2xdx`

`=int1/u(du)/2`

`=1/2int1/udu`

`=1/2ln|u|`

Substitute back `u=x^2+1`

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

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

Simplify and add a constant C to the solution,

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