Question

∫(1/sqrt(1-4x-x^2))dx

Answer 1

I got as far as completing the square

Answer 2

This requires a trig substitution.

Answer 3

Or using a table.

Answer 4

continueee...

Answer 5

What do you have when you completed the square?

Answer 6

1/Sqrt(1-(-4+4+4x+x^2)) = 1+4-(4++x+x^2) => 5-(x+2)^2

Answer 7

Good.

Answer 8

Now we need to let \(\sqrt{5}sin(\theta ) = x + 2\)

Answer 9

complete the square on \[1-4x-x^2\]

then use the substitution

\[\sqrt{5}u=x+2\]

Answer 10

Even easier, do a u-sub first. Let u=x+2
du=dx
Then you have:
\[\int\limits \frac{du}{\sqrt{5-u^2}}\]

Answer 11

Then use the substitution:
\[\sqrt5\sin(\phi)=u\]
\[du=\sqrt5 \cos(\phi)d \phi\]
\[\sqrt5\cos(\phi)=\sqrt{5-u^2}\]
Then you have:
\[\int\limits \frac{\sqrt5 \cos(\phi) d \phi}{\sqrt5 \cos(\phi)}=\int\limits d \phi=\phi+C\]
So solve for phi.
\[\sqrt5 \sin(\phi)=u \rightarrow \phi=\arcsin(\frac{u}{\sqrt5})\]
But u=x+2
\[\phi=\arcsin(\frac{x+2}{\sqrt5})\]
So your integral is:
\[\arcsin(\frac{x+2}{\sqrt5})+C\]