# integral of( 8x+10)/(sqrt(5-2x-x^2)

• Jan 27th 2013, 02:30 AM
Boo
integral of( 8x+10)/(sqrt(5-2x-x^2)
$\displaystyle \int \frac{8x-10}{\sqrt{5-2x-x^2}}$

how do I get rid of the sqrt in denominator?
Thanks!!
• Jan 27th 2013, 03:38 AM
ibdutt
Re: integral of( 8x+10)/(sqrt(5-2x-x^2)
• Jan 27th 2013, 04:28 AM
Prove It
Re: integral of( 8x+10)/(sqrt(5-2x-x^2)
Quote:

Originally Posted by Boo
$\displaystyle \int \frac{8x-10}{\sqrt{5-2x-x^2}}$
\displaystyle \displaystyle \begin{align*} \int{ \frac{8x - 10}{\sqrt{5 - 2x - x^2}} \,dx } &= \int{\frac{8x + 8}{\sqrt{5 - 2x - x^2}}\,dx} - \int{\frac{18}{\sqrt{-\left( x^2 + 2x - 5 \right)}}\,dx} \\ &= -4\int{\frac{-2x - 2}{\sqrt{5 - 2x - x^2}}} - \int{\frac{18}{\sqrt{-\left( x^2 + 2x + 1^2 - 1^2 - 5 \right)}}\,dx} \\ &= -4\int{ \frac{-2x-2}{\sqrt{5 - 2x - x^2}} \,dx } - \int{\frac{18}{\sqrt{-\left[ \left( x + 1 \right)^2 - 6 \right]}} \,dx} \\ &= -4\int{\frac{-2x-2}{\sqrt{5-2x-x^2}}\,dx} - \int{\frac{18}{\sqrt{6 - \left( x + 1 \right)^2}}\,dx} \end{align*}
In the first integral, make the substitution \displaystyle \displaystyle \begin{align*} u = 5-2x-x^2 \implies du = -2x-2\,dx \end{align*} and in the second, make the substitution \displaystyle \displaystyle \begin{align*} x + 1 = \sqrt{6}\sin{\theta} \implies dx = \sqrt{6}\cos{\theta}\,d\theta \end{align*}. See how you go :)