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

Rewrite the integral as ,

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

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

`=int(((x+1)(x-1))/(x-1)+1/(x-1))dx`

`=int(x+1+1/(x-1))dx`

apply the sum rule,

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

Apply the power rule and standard integralĀ `intdx/x=ln|x|`

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

Apply integral substitution `u=(x-1)` forĀ `int1/(x-1)dx`

`du=dx`

`int1/(x-1)dx=int(du)/u`

`=ln(u)`

substitute back`u=(x-1)`,

`=ln|x-1|`

So the final integration and adding a constant C to the solution yields,

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

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