Evaluate $\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$

Question

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

I completed the square:

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

And then substituted $\sqrt 2\sin θ = x-1$ which gives

$\int \dfrac{dθ}{\sqrt2 \sinθ + 2}$

But now I'm stuck. Can someone please help?

Answer 1

If you have a rational function involving trigonometric functions, try a Weierstrass substitution.

$$t=\tan\left(\frac {\theta}2\right)\qquad\mathrm d\theta=\frac {2}{1+t^2}\,\mathrm dt\qquad\sin\theta=\frac {2t}{1+t^2}$$

Hence

$$\begin{align*}\int\frac {\mathrm d\theta}{2+\sqrt{2}\sin\theta} & =\int\frac {\mathrm dt}{t^2+t\sqrt{2}+1}\\ & =\int\frac {\mathrm dt}{\left(t+\frac 1{\sqrt{2}}\right)^2+\frac 12}\end{align*}$$

Can you complete the rest?

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Another Approach

I also want to bring to light another possible substitution that, as far as I'm aware, isn't normally taught in a school setting (at least I didn't get taught this).

If you're daring enough and complacent with a large amount of algebraic manipulation up-front, then another possible solution would be to enforce an Euler Substitution of the Second Kind.

The substitution states that for an integral whose integrand is a purely rational function (i.e., no trigonometric terms)

$$\int R\left(x, \sqrt{ax^2+bx+c}\right)\,\mathrm dx$$

Where we impose the condition $c>0$, then an Euler substitution of the Second Kind gives

$$\sqrt{ax^2+bx+c}=xt+\sqrt{c}\qquad\qquad x=\frac {2t\sqrt{c}-b}{a-t^2}$$

In this case, since we have $c>0$, then our substitution is simply

$$\sqrt{1+2x-x^2}=xt+1\qquad\qquad x=\frac {2(1-t)}{1+t^2}$$

With this given substitution for $x$, we have that

$$x+1=\frac {3-2t+t^2}{1+t^2}\qquad\sqrt{1+2x-x^2}=\frac {1+2t-t^2}{1+t^2}\qquad\mathrm dx=\frac {2(t^2-2t-1)}{(1+t^2)^2}\,\mathrm dt$$

Right from the get-go, we can see that a lot of terms will cancel, which is the impressive nature of Euler Substitutions

$$\begin{align*}\int\frac {\mathrm dx}{(1+x)\sqrt{1+2x-x^2}} & =-2\int\frac {1+2t-t^2}{(1+t^2)^2}\frac {1+t^2}{t^2-2t+3}\frac {1+t^2}{1+2t-t^2}\,\mathrm dt\\ & =-2\int\frac {\mathrm dt}{(t-1)^2+2}\\ & =\sqrt{2}\arctan\left(\frac {1-t}{\sqrt2}\right)+C\end{align*}$$

Substituting our expression back in, then we get an impressive result

$$\int\frac {\mathrm dx}{(1+x)\sqrt{1+2x-x^2}}=\sqrt{2}\arctan\left(\frac {1+x-\sqrt{1+2x-x^2}}{x\sqrt2}\right)+C$$

Confirmed by Wolfram Alpha

Answer 2

Another method: Start with $x=\dfrac 1t$ so $dx=\dfrac{-1}{t^2}dt$. We have \begin{align}\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}&= \int \dfrac{\frac{-dt}{t^2}}{\left(1+\dfrac 1t\right)\sqrt{1+\dfrac2t-\dfrac {1}{t^2}}}\\&= \int \dfrac{-dt}{(1+t)\sqrt{t^2+2t-1}}.\end{align} Now use $t+1=u$ so $dt=du$. $$\int \frac{-du}{u\sqrt{u^2-2}}$$ which can be solved by putting $u^2-2=p$.

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EDIT: Another, maybe more useful, substitution would be $x+1=v$.

Answer 3

You may continue with $\theta=\frac\pi2+t$

\begin{align}\int \frac{1}{\sqrt2 \sin\theta+ 2}d\theta = &\int \frac{1}{\sqrt2 \cos t+ 2}dt= \int \frac{1}{2\sqrt2 \cos^2\frac t2+ 2-\sqrt2}dt\\ =&\int \frac{2d(\tan\frac t2)}{(2-\sqrt2)\tan^2\frac t2+(2+\sqrt2)}=\sqrt2\tan^{-1}\frac{\tan\frac t2}{\sqrt2+1}+C \end{align} Or, simply integrate as follows $$\int \frac{dx}{(1+x)\sqrt{1+2x-x^2}} = \int \frac{d(\frac{x}{1+x})}{\sqrt{1-2(\frac{x}{1+x})^2}}= \frac1{\sqrt2}\sin^{-1} \frac{\sqrt2x}{1+x}+C $$