Question

integral problem

Answer 1

\[\int\limits \frac{ \sqrt{x^2-3} }{ x }\] for \[x \le-\sqrt{3}\]

Answer 2

I forgot to type the dx but its there

Answer 3

please help

Answer 4

For this one use "u" substitution
u = x^2 -3
du = 2x dx

Answer 5

actually you will need to do a second substitution

u = w^2
du = 2w dw

Answer 6

\[\rightarrow \frac{1}{2} \int\limits \frac{\sqrt{u}}{u+3}du\]
\[\rightarrow \int\limits \frac{w^2}{w^2 +3} dw\]
oops i didn't think this through ... i guess we still need a trig sub as well
w = sqrt3 tan
dw = sqrt3 sec^2
\[\rightarrow \int\limits \frac{3 \tan^2 \theta}{3(\tan^2 \theta +1)} \sqrt{3} \sec^2 \theta\]
\[= \sqrt{3} \int\limits \tan^2 \theta\]

Answer 7

where does the \[x \le \sqrt{3}\] come in?