Evaluating a rational radical integral with a quadratic inside

Key takeaway: This integral becomes manageable after rewriting the quadratic under the square root in a shifted form and using a trigonometric substitution or a direct algebraic simplification.

Key idea

The integrand

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

looks complicated because it combines a rational factor with a square root of a quadratic. The first useful step is to complete the square inside the radical:

$3-2x-x^2 = 4-(x+1)^2.$

So the integral becomes

$\int \frac{dx}{(1-x)\sqrt{4-(x+1)^2}}.$

That form is designed for a trigonometric substitution, since $\sqrt{4-u^2}$ suggests $u=2\sin\theta$ or $u=2\cos\theta$.

Step-by-step substitution

Let

$x+1=2\sin\theta,$ so $dx=2\cos\theta\,d\theta,$ and $\sqrt{4-(x+1)^2}=\sqrt{4-4\sin^2\theta}=2\cos\theta.$

Also,

$1-x = 1-(2\sin\theta-1)=2(1-\sin\theta).$

Substitute these into the integral:

= \int \frac{1}{2(1-\sin\theta)}\,d\theta.$$ Now use the identity $$\frac{1}{1-\sin\theta}=\frac{1+\sin\theta}{\cos^2\theta}.$$ Then $$\int \frac{1}{2(1-\sin\theta)}\,d\theta = \frac12\int \left(\sec^2\theta+\sec\theta\tan\theta\right)d\theta.$$ Integrating gives $$\frac12\left(\tan\theta+\sec\theta\right)+C.$$ ## Convert back to x Since $x+1=2\sin\theta$, $$\sin\theta=\frac{x+1}{2}, \qquad \cos\theta=\frac{\sqrt{4-(x+1)^2}}{2}.$$ So $$\tan\theta=\frac{x+1}{\sqrt{4-(x+1)^2}}=\frac{x+1}{\sqrt{3-2x-x^2}},$$ and $$\sec\theta=\frac{2}{\sqrt{4-(x+1)^2}}=\frac{2}{\sqrt{3-2x-x^2}}.$$ Therefore the antiderivative is $$\frac12\left(\frac{x+1}{\sqrt{3-2x-x^2}}+\frac{2}{\sqrt{3-2x-x^2}}\right)+C =\frac{x+3}{2\sqrt{3-2x-x^2}}+C.$$ ## Common structure to remember Whenever a quadratic under a square root can be written as $a^2-u^2$, a trig substitution is often the cleanest route. Here the shifted variable $x+1$ was the crucial clue. The rational factor $1/(1-x)$ then simplified nicely after substitution, making the integral elementary. So the final answer is $$\boxed{\frac{x+3}{2\sqrt{3-2x-x^2}}+C}.$$ --- **Pitfalls the pros know** 👇 A common mistake is to start differentiating a guessed answer without first rewriting the quadratic inside the radical. If you miss the completion of the square, the substitution becomes much harder to spot. Another frequent error is handling $1-x$ incorrectly after substituting $x+1=2\sin\theta$; it becomes $2(1-\sin\theta)$, not $1-2\sin\theta$. Finally, be careful with the final back-substitution: both $\tan\theta$ and $\sec\theta$ must be expressed in terms of $\sqrt{3-2x-x^2}$, and dropping the factor of $2$ changes the answer. **What if the problem changes?** If the problem were changed to $$\int \frac{dx}{(x+1)\sqrt{3-2x-x^2}},$$ the same completion of the square would still give $4-(x+1)^2$ under the radical, but the rational factor would no longer simplify the same way. In that case, a substitution such as $x+1=2\sin\theta$ would turn the denominator into $2\sin\theta\cdot 2\cos\theta$, leading to a different trigonometric integral, often involving $\csc\theta$ or $\cot\theta$. If the interval or domain were altered so that $3-2x-x^2<0$, the real-valued antiderivative would no longer apply directly, and the problem would require a complex-analysis interpretation or a restricted real domain. `Tags`: completion of the square, trigonometric substitution, radical rationalization