Evaluate $\int \frac {x+3}{\sqrt{5-4x-2x^2}} dx$ Question:
The question is about evaluating the following indefinite integral:-
$$\int \frac {x+3}{\sqrt{5-4x-2x^2}}dx$$
My Work:
We know the numerator can be expressed in the form of :
$$ x+3 = A \frac {d}{dx}(5-4x-2x^2) + B$$
So we get ,
$$ x+3 = A(-4-4x) +B $$
Equating $x$ and $3$ we get,
$$ A = \frac {1}{4} , B={4} $$
So integrating,
$$ \int \frac {1}{4} ({{-4-4x}})\frac{1}{\sqrt{5-4x-2x^2}}dx + \int \frac {4}{\sqrt{5-4x-2x^2}}dx$$
Taking the constants out,
Assume $5-4x-2x^2 = t$
Differentiating it we get,
$$(-4-4x)dx= dt$$
By completing the square method we get,
$$5-4x-2x^2 = (2)(\frac {5}{2}-2x-x^2)$$
So doing the calculation we get ,
$$(2)[{(x+1)^2-(\sqrt \frac{7}{2})^2}$$
Now again integrating all the values obtained,
$$\frac {1}{4} \int \frac {dt}{\sqrt {t}} - \frac {4}{\sqrt 2} \int \frac {dx}{(x+1)^2-(\sqrt \frac{7}{2})^2}$$
Taking the minus sign out,
$$\frac{1}{2}\sqrt{t}dt+\frac{4}{\sqrt{2}}\int_{ }^{ }\frac{dx}{\left(\sqrt{\frac{7}{2}}\right)^{2}-(x+1)^{2}}$$
$$\frac{1}{2}\sqrt{5-4x-2x^{2}}dt+\frac{4}{\sqrt{2}}\sin^{-1}\left(\frac{x+1}{\left(\sqrt{\frac{7}{2}}\right)^{2}}\right)$$
I cannot integrate it further any help or advice will be very much appreciated

Another method using Completing the square and Integration by substitution.
We have,
$$\begin{align}I &= \int \dfrac{x + 3}{\sqrt{5 - 4x - 2x^2}} dx\\& = \int \frac{x+3}{\sqrt{5 - 2(x^2 +2x)}}dx\\& = \int \frac{x+3}{\sqrt{5 - 2[(x + 1)^2 - 1]}}dx\\& = \int \frac{x+3}{\sqrt{7 - 2(x + 1)^2}}dx\end{align}$$
Let $(x+1) = \sqrt{\frac{7}{2}} \sin\theta \implies dx = \sqrt{\frac72}\cos\theta\ d\theta.$
So we have,
$$\begin{align}
I& = \int \frac{\left(\sqrt{\frac{7}{2}} \sin \theta + 2\right)}{\sqrt{7 - 2\cdot \frac72\sin^2\theta}}\cdot \sqrt{\frac72}\cos\theta\ d\theta
\\& = \frac1{\sqrt2}\int \frac{\left(\sqrt{\frac{7}{2}} \sin \theta + 2\right)}{\sqrt{7 - 7\sin^2\theta}}\cdot \sqrt{7}\cos\theta\ d\theta
\\& = \frac1{\sqrt2}\int \frac{\left(\sqrt{\frac{7}{2}} \sin \theta + 2\right)}{\sqrt{7}\cos\theta}\cdot \sqrt{7}\cos\theta\ d\theta
\\& = \frac1{\sqrt2}\int \sqrt{\frac{7}{2}} \sin \theta + 2\ d\theta
\\& = -\frac{\sqrt7}{2} \cos \theta +\sqrt{2}\theta + C
\end{align}$$
Consider our substitution,
$$ \sqrt{\frac{2}{7}} (x+1) =\sin\theta \implies \cos\theta = \sqrt{1 -\sin^2\theta} = \sqrt{1 - \frac27 (x+1)^2}.$$
Thus, our integral becomes,
$$ I = -\frac{\sqrt7}{2} \cos \theta +\sqrt{2}\theta + C$$
$$ -\frac{\sqrt7}{2} \cdot\sqrt{1 - \frac27 (x+1)^2} +\sqrt{2}\cdot \sin^{-1}\left(\sqrt{\frac{2}{7}} (x+1)\right)+ C$$
$$ \color{blue}{\boxed{-\frac{1}{2}\sqrt{-2x^{2}-4x+5}+\sqrt{2}\cdot \sin^{-1}\left(\sqrt{\frac{2}{7}} (x+1)\right)+ C}}$$
This is the final answer.
