Im trying to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $.

I've tried integration by parts, and any resonable substitution that came to my mind. Neither seem to work. I tried to use integral online calculator, the result was too complicated, and I'm pretty sure there is a resonable way to solve it, as we had this question in an exam.

Thanks in advance.

  • $\begingroup$ Hint: $(2x-\sqrt{4x^{2}-x+1})(2x+\sqrt{4x^{2}-x+1})=x-1$ $\endgroup$ – Eminem Aug 15 '20 at 12:21
  • $\begingroup$ It's not an impossible integral but the solution is not particularly simple for an exam question. $\endgroup$ – Peter Foreman Aug 15 '20 at 12:22
  • $\begingroup$ First thing I tried was the hint suggested in the first comment, it did not lead me to the solution. Can someone point out a way that leads to the solution ? (just name the steps and I'll do it myself) $\endgroup$ – FreeZe Aug 15 '20 at 12:26


Then,$$\sqrt{4x^2-x+1}=2x+t \;\;\Longrightarrow\;\;4x^2-x+1=(2x+t)^2 \;\;\Longrightarrow\;\; x=\frac{1-t^2}{4t+1}\,.$$

And $$dx = -\frac{4t^2+2t+4}{(4t+1)^2}dt\,.$$



$$\begin{align} & =-\int\frac{4t^2+2t+4}{(4t+1)(t+4)}dt\\ &=-\int\frac{4t^2+2t+4}{4t^2+17t+4}dt\\ &=-\int\left(1-\frac{15t}{4t^2+17t+4}\right)dt\\ &=-t+\int\frac{15t}{(4t+1)(t+4)}dt\,. \end{align}$$

Splitting into partial fractions,

$$\frac{15t}{(4t+1)(t+4)} = \frac{4}{t+4}-\frac{1}{4t+1}\,,$$

We get,


so, therefore,

$$\Rightarrow \bbox[5px,border:2px solid red]{I=-t+4\ln(t+4)-\frac{1}{4}\ln(4t+1)+C\,,}$$

On rewriting we get,

$$\Rightarrow \bbox[5px,border:2px solid red]{\begin{align}I\;=\;&2x-\sqrt{4x^2-x+1}+4\ln(\sqrt{4x^2-x+1}-2x+4)\\ &-\frac{1}{4}\ln(4\sqrt{4x^2-x+1}-8x+1)+C\,. \end{align}}$$


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