# Thread: Final integrals

1. ## Final integrals

I want to compare my answers to your's?
I will be glad if someone will solve them?

2. If you only want to compare solutions, I'm giving my answers for now:

4) 1/6

5) $\frac{2\cdot5^{{\sqrt{1 + 3\,x}}}}{3\cdot{\log (5)}^2}\left({\sqrt{1 + 3\,x}}\,\log (5)-1 \right)$

10) doesn't converge

11) $\frac{\pi}{8 \cdot \sqrt{5}}$

If you have something else, I can post more details.

3. Hello, sbsite!

This is what I got for #4 . . .

$4)\;\;\int^2_1\frac{dx}{(x^2 - 2x + 4)^{\frac{3}{2}}}$

The denominator has: . $x^2 - 2x + 1 + 3\:=\:(x-1)^2 + 3$

Let $x - 1 \:=\:\sqrt{3}\tan\theta\quad\Rightarrow\quad dx \:=\:\sqrt{3}\sec^2\theta\,d\theta$

When $x = 1\!:\;\;\sqrt{3}\tan\theta = 0\quad\Rightarrow\quad\theta = 0$
When $x = 2\!:\;\;\sqrt{3}\tan\theta = 1\quad\Rightarrow\quad\tan\theta = \frac{1}{\sqrt{3}}\quad\Rightarrow\quad\theta = \frac{\pi}{6}$

Substitute: . $\int^{\frac{\pi}{6}}_0 \frac{\sqrt{3}\sec^2\theta\,d\theta}{(3\sec^2\thet a)^{\frac{3}{2}}} \;=\;\int^{\frac{\pi}{6}}_0\frac{\sqrt{3}\sec^2\th eta\,d\theta}{3\sqrt{3}\sec^3\theta} \;=\;\frac{1}{3}\int^{\frac{\pi}{6}}_0\frac{d\thet a}{\sec\theta}$ $= \;\frac{1}{3}\int^{\frac{\pi}{6}}_0\cos\theta\,d\t heta$

And we have: . $\frac{1}{3}\sin\theta\,\bigg|^{\frac{\pi}{6}}_0 \;=\;\frac{1}{3}\sin\frac{\pi}{6} - \frac{1}{3}\sin0 \;=\;
\frac{1}{3}\left(\frac{1}{2}\right) - \frac{1}{3}(0) \;=\;\boxed{\frac{1}{6}}$