Math Help - Integration

1. Integration

∫ (3x-1)/(x^2-2x+10)^0.5 dx

How can I integrate this?

2. Hi

$\frac{3x-1}{\sqrt{x^2-2x+10}}=\frac{3x-1}{\sqrt{x^2-2x+1+9}}=(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}$

Transforming the fraction to make appear $\mathrm{arcsinh}'u=\frac{1}{\sqrt{u^2+1}}$ may lead to the result. Any idea on how it can be achieved ?

3. Originally Posted by flyingsquirrel
Any idea on how it can be achieved ?
Actually, no. Can you tell me, please?

4. Originally Posted by flyingsquirrel
Hi

$\frac{3x-1}{\sqrt{x^2-2x+10}}=\frac{3x-1}{\sqrt{x^2-2x+1+9}}=(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}$

Transforming the fraction to make appear $\mathrm{arccosh}'u=\frac{1}{\sqrt{u^2+1}}$ may lead to the result. Any idea on how it can be achieved ?
I know that, 1/√(1+x^2) = arsinh x.

But I don’t know what is the next step of $(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}$

5. Originally Posted by geton
I know that, 1/√(1+x^2) = arsinh x.

But I don’t know what is the next step of $(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}$
We want the fraction to be $\frac{1}{\sqrt{u^2+1}}$ and the square is already present. To make appear 1, let's factor by 9 inside the square root :

$(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}=(3x-1)\cdot\frac{1}{\sqrt{9\left(\frac{(x-1)^2}{9}+1\right)}}=(3x-1)\cdot\frac{1}{3\sqrt{\left(\frac{x-1}{3}\right)^2+1}}$

Hence $\frac{3x-1}{\sqrt{x^2-2x+10}}=(3x-1)\cdot\frac{1}{3\sqrt{\left(\frac{x-1}{3}\right)^2+1}}$

Substituting $u=\frac{x-1}{3}$ makes appear $\mathrm{arcsinh}'u$... but it is multiplied by a polynomial. How can one get rid off it ?

6. Originally Posted by geton
∫ (3x-1)/(x^2-2x+10)^0.5 dx

How can I integrate this?
$\frac{3x - 1}{\sqrt{x^2 - 2x + 10}} = \frac{3 \left( x - \frac{1}{3} \right)}{\sqrt{x^2 - 2x + 10}} = \frac{\frac{3}{2} \left( 2x - \frac{2}{3} \right)}{\sqrt{x^2 - 2x + 10}}$

$= \frac{\frac{3}{2} \left( [2x - 2] + 2 - \frac{2}{3} \right)}{\sqrt{x^2 - 2x + 10}} = \frac{3}{2} \, \frac{2x - 2}{\sqrt{x^2 - 2x + 10}} + \frac{3}{2} \, \frac{2 - \frac{2}{3}}{\sqrt{x^2 - 2x + 10}}$

$= \frac{3}{2} \, \frac{2x - 2}{\sqrt{x^2 - 2x + 10}} + \frac{2}{\sqrt{x^2 - 2x + 10}}$.

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.

7. Thank you so much mr fantastic for help. And sorry for late reply, because of study pressure.