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.

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, 2020 at 12:21
  • $\begingroup$ It's not an impossible integral but the solution is not particularly simple for an exam question. $\endgroup$ Aug 15, 2020 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, 2020 at 12:26

3 Answers 3



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,

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


Complete the square to apply the trigonometric substitution $$2x-\frac14=\frac{\sqrt{15}}4\cot t, \>\>\>\text{or}\>\>\>g(x)=\sqrt{4x^2-x+1}=\frac{\sqrt{15}}4\csc t $$ Then \begin{align} I=& \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}\ dx \\ =&- \frac12\int \frac{\csc t}{1+\cos t + \frac1{\sqrt{15}}\sin t}\ dt\\ =& -\frac14 \int \csc t +\frac{\sqrt{15}}{1+\cos t}-\frac{16(\csc t -\cot t)}{1-\cos t +\frac1{\sqrt{15}}\sin t}\ dt \end{align} where $\int \csc t\ dt= -\tanh^{-1}(\cos t)$

\begin{align} &\int \frac{1}{1+\cos t}\ dt =\csc t -\cot t\\ &\int \frac{\csc t -\cot t}{1-\cos t +\frac1{\sqrt{15}}\sin t}\ dt= \ln (\sqrt{15}+\csc t -\cot t) \end{align} After back-substitution $$I= 2x-g(x)+ \frac14\tanh^{-1}\frac {2x-\frac14}{g(x)} +4\ln\left(4+g(x)-2x\right) $$


$$ \begin{aligned} I=& \int \frac{2 x-\sqrt{4 x^2-x+1}}{x-1} d x \\ = & 2 \underbrace{\int \frac{x}{x-1} d x}_{=2x+2\ln |x-1|} - \underbrace{\int \frac{\sqrt{4 x^2-x+1}}{x-1} dx}_{J} \end{aligned} $$ Letting $z=\frac{1}{x-1} $ transforms the integral into $$ J=-\int \sqrt{4 z^2+7 z+4} \,d z= -\int \sqrt{\left(2 z+\frac{7}{4}\right)^2+\frac{15}{16}}\, d z $$ Using the hyperbolic substitution $2z+\frac{7}{4}=\frac{\sqrt{15}}{4} \sinh \theta$, we have

$$ J=-\frac{1}{64}\left[4 \sqrt{4 z^2+7 z+4}\,(8 z+7)+15 \sinh ^{-1}\left(\frac{8 z+7}{\sqrt{15}}\right)\right]+C $$ where $z=\frac{1}{x-1} $. $$I= 2x+2\ln |x-1| -\frac{1}{16}\left[4 \sqrt{\frac{4 x^2-x+1}{(x-1)^2}}\left(\frac{7 x+1}{x-1}\right)+15 \sinh ^{-1}\left(\frac{7 x+1}{\sqrt{15}(x-1)}\right)\right]+C $$


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