Question

integrate (4x+8)/sqrt(4x^2+8x+3)

Answer 1

Try a trig substitution. First, notice that for the denominator (while ignoring the square root for now), you can complete the square:

\[\begin{align} 4x^2+8x+3 &=4\left( x^2 + 2 + \frac{3}{4}\right)\\&=4\left( x^2 + 2 +1-1+\frac{3}{4}\right)\\&=4\left[(x+1)^2-\frac{1}{4}\right] \\&=4(x+1)^2 - 1 =2^2(x+1)^2-1\end{align} \]

So, \[\int\frac{4x+8}{\sqrt{4(x+1)^2-1}}\, dx \]
Now, you can use trig substitution. Notice that the denominator looks like it has the form \(\sqrt{x^2-a^2}\) but in fact your "x" is (x+1), with a factor of 4 in front, and "a" is -1, suggesting to use the substitution \[(x+1) = \frac{1}{2}\sec \theta \implies x = \frac{1}{2}\sec \theta - 1 \\
\\ \, \\dx=\frac{1}{2}\sec \theta \tan \theta \, d\theta\] Substituting all this gives:
\[ \int\frac{4 \left(\frac{1}{2}\sec \theta-1 \right)+8}{\sqrt{4\left(\frac{1}{2}\sec \theta\right)^2-1}} \left( \frac{1}{2}\sec \theta \tan \theta \right) \, d \theta \\ \, \\=\int\frac{2 \sec \theta -4+8}{\sqrt{4 \left( \frac{1}{4}\sec^2 \theta \right)-1 }}\left( \frac{1}{2}\sec \theta \tan \theta \right) \, d \theta \\ \, \\ =\int\frac{2 \sec \theta +4}{\sqrt{\sec^2 \theta -1 }} \left( \frac{1}{2}\sec \theta \tan \theta \right) \, d \theta \\ \, \\ =\int\frac{2\sec \theta + 4}{\tan \theta} \left( \frac{1}{2}\sec \theta \tan \theta \right) \, d \theta, \text{ since } \sec^2 \theta -1 = \tan^2\theta \\ \, \\=\int(\sec \theta + 2)\sec \theta \ d\theta \\ \, \\ =\int (\color{red}{\sec^2 \theta}+ \color{blue}{2 \sec \theta}) \, d \theta \\ \, \\= \color{red}{\tan \theta} + \color{blue}{2 \ln|\sec \theta + \tan \theta |}+ C\]

Now to go back to x variables.. notice from above that \(\sec \theta = 2(x+1) \) and in the denominator, our \(\tan \theta\) was the result of the simplification from \(\sqrt{4(x+1)^2-1}\), hence :
\[ =\color{red}{\sqrt{4(x+1)^2-1}} + \color{blue}{2\ln\left|\sqrt{4(x+1)^2 - 1}+2(x+1) \right|}\] and you can "simplify" that a bit if you which by expanding the terms under the square root

Answer 2

+ C..

Answer 3

A somewhat different approach...
\[\int\frac{4x+8}{\sqrt{4x^2+8x+3}}~ dx=\frac{1}{2}\int\frac{8x+8}{\sqrt{4x^2+8x+3}}~ dx+\int\frac{4}{\sqrt{4(x+1)^2-1}}~ dx\]
Then the first integral is just a matter of substituting \(u=4x^2+8x+3\) so that \(du=8x+8\). The second integral still uses a trig sub though.

Answer 4

^I tried that method first too :) But realized I had to do a trig sub too for the last one.