Integral of [x+3÷(5-x²-4x)½]dx?

Apr 21, 2018

$- \sqrt{5 - {x}^{2} - 4 x} + {\sin}^{- 1} \left(\frac{x + 2}{3}\right) + C$

Explanation:

5-x²-4x = 9 - (x + 2)^2

So the integral is:

$\int \setminus \frac{x + 3}{\sqrt{9 - {\left(x + 2\right)}^{2}}} \setminus \mathrm{dx}$

$\int \setminus \frac{x + 2}{\sqrt{9 - {\left(x + 2\right)}^{2}}} + \setminus \frac{1}{\sqrt{9 - {\left(x + 2\right)}^{2}}} \setminus \mathrm{dx}$

This is handy on 2 counts.

For the first part: note that: $\frac{d}{\mathrm{dx}} \left(\sqrt{9 - {\left(x + 2\right)}^{2}}\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{9 - {\left(x + 2\right)}^{2}}} \cdot - 2 \left(x + 2\right) = - \frac{x + 2}{\sqrt{9 - {\left(x + 2\right)}^{2}}}$

And so:

$\int \setminus \frac{x + 2}{\sqrt{9 - {\left(x + 2\right)}^{2}}} = - \int \setminus d \left(\sqrt{9 - {\left(x + 2\right)}^{2}}\right) = - \sqrt{9 - {\left(x + 2\right)}^{2}} + C = - \sqrt{5 - {x}^{2} - 4 x} + C$

For the second part, if we let $x + 2 = 3 \sin \setminus \omega$, then we have:

$\int \setminus \setminus \frac{1}{\sqrt{9 - 9 {\sin}^{2} \setminus \omega}} \setminus \setminus d \left(3 \sin \setminus \omega - 2\right)$

$\int \setminus \setminus \frac{3 \cos \omega}{3 \cos \omega} \setminus \setminus d \omega$

$= \omega + C$

$= {\sin}^{- 1} \left(\frac{x + 2}{3}\right) + C$

Combining these:

$I = - \sqrt{5 - {x}^{2} - 4 x} + {\sin}^{- 1} \left(\frac{x + 2}{3}\right) + C$