anonymous 5 years ago What is the value of ∫[1,2,1/[(x+2)^2√((x+2)^2+16)],]?

1. anonymous

Okay, I have your answer, but this website is giving me grief re. posting so I can't write huge slabs.

2. anonymous

Let$w^2=(x+2)^2+16$then$2w.dw=2(x+2).dx \rightarrow dx = \frac{w}{x+2}dw=\frac{w}{\sqrt{w^2-16}}dw$

3. anonymous

so your integral becomes,$\int\limits_{w_1}^{w_2}\frac{w}{\sqrt{w^2-16}}\frac{1}{w(w^2-16)} dw=\int\limits_{w_1}^{w_2}\frac{1}{(w^2-16)^{3/2}} dw$

4. anonymous

Now let w=cosh(theta). Then dw=sinh(theta) d(theta) and w^2-16 = 16sinh^2(theta). Substitute into the integral:$\int\limits_{\theta_1}^{\theta_2}\frac{4\sinh \theta}{(16\sinh^2 \theta)^{3/2}}d \theta=\frac{1}{16}\int\limits_{\theta_1}^{\theta_2}\frac{d \theta }{\sinh^2 \theta}=\frac{1}{16}\int\limits_{\theta_1}^{\theta_2}cosech^2 \theta d \theta$

5. anonymous

$=-\frac{1}{16}\coth \theta|^{\theta_2}_{\theta_1}$Using the fact that coth(theta)= cosh(theta)/sinh(theta), you have$\coth \theta = \frac{\cosh \theta}{\sinh \theta}=\frac{\cosh \theta }{\sqrt{\cosh^2 \theta -1}}=\frac{w}{\sqrt{w^2-16}}$Using the definition of the sub. for w, and the result for the integral, $\int\limits_{}{}\frac{1}{(x+2)^2\sqrt{(x+2)^2+16}}dx=-\frac{1}{16}\frac{w}{\sqrt{w^2-16}}$

6. anonymous

$-\frac{1}{16}\frac{\sqrt{(x+2)^2+16}}{x+2}|^{2}_{1}$

7. anonymous

phew...

8. anonymous

Thank you so much!!!!!!!!!

9. anonymous

You're welcome :) Become a fan if you haven't already!