`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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