How to solve $\int{\frac{1}{\sqrt{3-2x-x^2}}\,dx}$?

Question

$$\int{\frac{1}{\sqrt{3-2x-x^2}}\,dx}$$

I tried to do it by substitution with no sucess. Anyone can solve it?

Answer 1

The thing you should think of instantly when you see a thing like that is that completing the square is the standard method in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial with no first-degree term.

In this case $3-2x-x^2 = 4 - (x+1)^2 = 4-u^2$. Now you have a constant term, $4$, and a quadratic term, $-u^2$.

Next: Do you have \begin{align} \sqrt{a^2-u^2} \\[6pt] \text{or }\sqrt{u^2-a^2} \\[6pt] \text{or }\sqrt{u^2+a^2} & {}\quad \text{?} \end{align} Each yields to a different trigonometric subsitution.

Answer 2

HINT: Complete the square $$3 - 2x - x^2 = 4-1-2x-x^2 =2^2 - (x+1)^2$$

Then put $(x+1) = 2\sin \theta$.

Answer 3

Completing the square and using the substitution $u=x+1$ gives us

\begin{equation*} \int\frac{1}{\sqrt{4-(x+1)^2}}dx=\int\frac{1}{\sqrt{4-u^2}}du=\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{u^2}{4}}}. \end{equation*}

Using the substitution $s=\frac{u}{2}$ gives

\begin{equation*} \int\frac{1}{\sqrt{1-s^2}}=\sin^{-1}(s)+C. \end{equation*}

Substituting back $s=\frac{u}{2}$ and $u=x+1$ gives the result.