# $\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$ Integrate

$$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$$

What I first did is I tried to complete the square for the numerator

$\sqrt{4(x^2-2x+3/4)} = \sqrt{4(x^2-2x+1-1/4)} = \sqrt{4(x-1)^2 - 1}$

Now I did trig-substitution:

$(x-1) = \frac{1}{2}\sec(\theta)$

$dx = \frac{1}{2}\sec(\theta)\tan(\theta)d\theta$

$$=\int \frac{\tan^2(\theta)\sec(\theta)}{\frac{1}{2}\sec(\theta) }d\theta = 2\tan(\theta) - 2\theta + C = 2\frac{\sqrt{4x^2-8x+3}}{2} - 2\sec^{-1}\left(\frac{x-1}{2}\right) + C$$

• You made a good attempt. Then why not continue by assuming $x-1=u$? – Juniven Apr 17 '17 at 6:43
• ohhhhhhhhhhhhh not use that thx i get it – Tinler Apr 17 '17 at 6:44
• Then the integral becomes $$\int\frac{\sqrt{4u^2-1}}{u}du$$ – Juniven Apr 17 '17 at 6:44
• Like that? I edited – Tinler Apr 17 '17 at 6:47
• set $t=2\cosh(y)$ – Dr. Sonnhard Graubner Apr 17 '17 at 6:51

If you substitute $u=\sqrt{4x^2-8x+3}$ you obtain $du=\dfrac{8x-8}{2\sqrt{4x^2-8x+3}}dx$. So : $$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx={\displaystyle\int}\dfrac{u^2}{u^2+1}\,du$$

Perform long division and you can finish easily.

I think your last step is wrong.

$2\tan\theta - 2\theta + c$

$= 2 \sqrt {\sec^2 \theta - 1} - 2 \theta + c$

As we have,

$\sec \theta = 2(x - 1)$

$\theta = \sec^{-1}2(x - 1)$

Put these values above,

$= 2 \sqrt {4x^2-8x+3} - 2 \sec^{-1}2(x - 1)+ c$