∫ (3x-1)/(x^2-2x+10)^0.5 dx
How can I integrate this?
Hi
$\displaystyle \frac{3x-1}{\sqrt{x^2-2x+10}}=\frac{3x-1}{\sqrt{x^2-2x+1+9}}=(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}$
Transforming the fraction to make appear $\displaystyle \mathrm{arcsinh}'u=\frac{1}{\sqrt{u^2+1}}$ may lead to the result. Any idea on how it can be achieved ?
We want the fraction to be $\displaystyle \frac{1}{\sqrt{u^2+1}}$ and the square is already present. To make appear 1, let's factor by 9 inside the square root :
$\displaystyle (3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}=(3x-1)\cdot\frac{1}{\sqrt{9\left(\frac{(x-1)^2}{9}+1\right)}}=(3x-1)\cdot\frac{1}{3\sqrt{\left(\frac{x-1}{3}\right)^2+1}}$
Hence $\displaystyle \frac{3x-1}{\sqrt{x^2-2x+10}}=(3x-1)\cdot\frac{1}{3\sqrt{\left(\frac{x-1}{3}\right)^2+1}}$
Substituting $\displaystyle u=\frac{x-1}{3}$ makes appear $\displaystyle \mathrm{arcsinh}'u$... but it is multiplied by a polynomial. How can one get rid off it ?
$\displaystyle \frac{3x - 1}{\sqrt{x^2 - 2x + 10}} = \frac{3 \left( x - \frac{1}{3} \right)}{\sqrt{x^2 - 2x + 10}} = \frac{\frac{3}{2} \left( 2x - \frac{2}{3} \right)}{\sqrt{x^2 - 2x + 10}}$
$\displaystyle = \frac{\frac{3}{2} \left( [2x - 2] + 2 - \frac{2}{3} \right)}{\sqrt{x^2 - 2x + 10}} = \frac{3}{2} \, \frac{2x - 2}{\sqrt{x^2 - 2x + 10}} + \frac{3}{2} \, \frac{2 - \frac{2}{3}}{\sqrt{x^2 - 2x + 10}} $
$\displaystyle = \frac{3}{2} \, \frac{2x - 2}{\sqrt{x^2 - 2x + 10}} + \frac{2}{\sqrt{x^2 - 2x + 10}}$.
Integral of the first term is done using an obvious substitution. Integral of the second term is done by completing the square inside the square root etc.