Evaluate $\int \frac{x^2}{x-1} \,dx$

(A) $2x^2+x+\ln|x-1|+C$

(B) $\frac{x^2}{2}+x+\ln|x+1|+C$

(C) $\frac{x^2}{2}+x+\ln|x-1|+C$

(D) $x^2+x+\ln|x-1|+C$

My attempt :

Let $u=x-1$, so : $du=dx$ and $u+1$

$\int \frac{(u+1)^2}{u}\,du\\ =\int u +2+\frac{1}{u}\,du\\ =\frac{u^2}{2}+2u+\ln|u|+C\\ =\frac{(x-1)^2}{2}+2(x-1)+\ln|x-1|+C$

Simplify :


It's not on the option.

  • 3
    $\begingroup$ Your answer is $\frac{x^2}2+x+\ln|x-1|+(C-\frac32)$. Can you resolve the matter from here? $\endgroup$
    – David
    Feb 1, 2019 at 3:12
  • $\begingroup$ @David so, is it C? $\endgroup$
    – love you
    Feb 1, 2019 at 3:15
  • $\begingroup$ yea you can break up the $\frac{x^2-3}{2}$ $\endgroup$
    – user29418
    Feb 1, 2019 at 3:16

3 Answers 3


You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer $$\frac{x^2-3}{2}+x+\ln|x-1|+C_1$$ equal to option (C) and simplify. You'll get $C-C_1=-\frac32$, which is fine since the difference is constant.

Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $\frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.

As a side note, it isn't really correct to say that $\int\frac1x\ dx=\ln|x|+C$, because of the discontinuity of $\frac1x$ at $0$. It's a bit more nuanced than that: $$ \int\frac1x\ dx = \begin{cases} \ln|x| + C_1, & \text{if $x > 0$} \\ \ln|x| + C_2, & \text{if $x < 0$} \end{cases}$$

So in that sense, even the given answers are not entirely right.


Your answer is right except for the constant I changed. $$\frac{x^2-3}{2}+x+\ln|x-1|+C_1 = \frac{x^2}{2}+x+\ln|x-1|+C_1-\frac{3}{2} = \frac{x^2}{2}+x+\ln|x-1|+C$$


After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:

$$ \begin{align} \int \frac{x^2}{x-1} \,dx &=\int \frac{x^2-1+1}{x-1} \,dx\\ &=\int \left(\frac{x^2-1}{x-1}+\frac{1}{x-1}\right) \,dx\\ &=\int \frac{(x-1)(x+1)}{x-1}\,dx+\int\frac{1}{x-1} \,dx\\ &=\int \left(x+1\right)\,dx+\int\frac{1}{x-1}\frac{d}{dx}\left(x-1\right) \,dx\\ &=\frac{x^2}{2}+x+\int\frac{1}{x-1}\,d\left(x-1\right)\\ &=\frac{x^2}{2}+x+\ln{\left|x-1\right|}+C.\\ \end{align} $$

  • 1
    $\begingroup$ The OP has already done all of this; the question concerns the constant at the end. $\endgroup$
    – Théophile
    Feb 1, 2019 at 20:22

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