Find $\int \frac{dx}{\sqrt{(1-2x-x^2)}}$ completing square.

Question

Find $\int \frac{dx}{\sqrt{(1-2x-x^2)}}$ completing square.

$$=\int \frac{dx}{\sqrt{-(x+1)^2+2}}$$

I know that $\int \frac{dx}{\sqrt{x^2+a}}=\ln|x+\sqrt{x^2+a}|+C$

but here we have $-(x+1)^2$.

So with method shown in the image I tried to take out formula for $-(x+1)^2$

but in the image $t^2+a$'s are cancelling which didn't happen in the example that I tried.

Answer 1

Let $1+x=\sqrt{2}t,$ $dx=\sqrt{2}dt$ so we have

$$\int \frac{dx}{\sqrt{-(x+1)^2+2}}=\int\frac{\sqrt{2}dt}{\sqrt{2-2t^2}}=\int\frac{1}{\sqrt{1-t^2}}dt$$

Now let $t=\sin(v), dt=\cos(v)dv$ which gives

$$\int\frac{1}{\sqrt{1-t^2}}dt=\int\frac{\cos(v)}{\sqrt{1-\sin^2(v)}}dv=\int dv=...$$

Note: substituting $1+x=\sqrt{2}\sin(v)$ gets you directly to the last integral.

Answer 2

There's a couple of integrals you may want to know the methods to as well as the one you stated: $$\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+C$$ $$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+C$$ $$\int\frac{dx}{\sqrt{a^2+x^2}}=\operatorname{arsinh}\frac xa+C$$ $$\int\frac{dx}{a^2-x^2}=\frac1a\operatorname{artanh}\frac xa+C$$ just notice what form yours is in and one of these may be easier :)