What is the easiest way to integrate $y=\frac {x+4}{\sqrt{-x^2-2x+3}}$?

Question

What is the easiest way to integrate $y=\frac{x+4}{\sqrt{-x^2-2x+3}}$ ? I tried to integrate it by making numerator in form: $-2x-2$ and then pulling it under differential, but the result drastically differed from Mathematica output.

Answer 1

Since $-x^2-2x+3=-(x+3)(x-1)$, both roots $x_1=-3,x_2=1$ are real. As such it is known that e.g. the Euler substitution

\begin{align} \sqrt{-x^{2}-2x+3}&=t(x-x_2)=t(x-1)\\[2ex] &\Leftrightarrow x=\frac{t^{2}-3}{t^{2}+1}=1-\frac{4 }{1+t^{2}},\quad dx=\frac{8t}{\left( 1+t^{2}\right) ^{2}}dt, \\[2ex] & \Leftrightarrow\sqrt{-x^{2}-2x+3}=-\frac{4t}{1+t^{2}} \end{align}

will turn the original integral into an integral of a rational function in $t$ which can be integrated by the method of partial fractions:

\begin{eqnarray*} I &=&\int \frac{x+4}{\sqrt{-x^{2}-2x+3}}dx = \int\frac {1-{\dfrac {4}{1+t^2}+4}}{-\dfrac {4t}{1+t^2}}\frac {8t}{(1+t^2)^2}\, dt\\[2ex] &=&\int -2\frac{1+5t^{2}}{\left( 1+t^{2}\right) ^{2}}dt=\int \frac{8}{\left( 1+t^{2}\right) ^{2}}- \frac{10}{1+t^{2}}dt\\[2ex] &=&\int \frac{8}{\left( 1+t^{2}\right) ^{2}}dt-\int \frac{10}{1+t^{2}}dt,\quad \text{(see note below) }\\[2ex] &=&\left(\frac{4t}{1+t^{2}}+4\arctan t\right)-10\arctan t+C \\[2ex] &=&\frac{4t}{1+t^{2}}-6\arctan t +C\\[2ex] &=&-\sqrt{-x^{2}-2x+3}-6\arctan \frac{\sqrt{-x^{2}-2x+3}}{x-1}+C. \end{eqnarray*}

Note.

• By the method explained here we can reduce the integration of the function $f(t)=\frac{1}{\left( 1+t^{2}\right) ^{2}}$ to the integration of $\frac{1}{1+t^{2}}$. We start by adding and subtracting $t^2$ in the numerator. The first integral is a standard integral and the second one is integrable by parts: $$\begin{eqnarray*} \int \frac{1}{\left( 1+t^{2}\right) ^{2}}dt &=&\int \frac{1}{1+t^{2}}dx-\int \frac{t^{2}}{\left( 1+t^{2}\right) ^{2}}dt \\ &=&\arctan t-\int t\frac{t}{\left( 1+t^{2}\right) ^{2}}dt, \end{eqnarray*}$$ and $$\begin{eqnarray*} \int t\frac{t}{\left( 1+t^{2}\right) ^{2}}dt &=&t\left( -\frac{1}{2\left( 1+t^{2}\right) }\right) +\int \frac{1}{2\left( 1+t^{2}\right) }dt \\ &=&-\frac{t}{2\left( 1+t^{2}\right) }+\frac{1}{2}\arctan t. \end{eqnarray*}$$ We thus get $$\int \frac{1}{\left( 1+t^{2}\right) ^{2}}dt =\frac{t}{2\left( 1+t^{2}\right) }+\frac{1}{2}\arctan t,$$ $$\int \frac{8}{\left( 1+t^{2}\right) ^{2}}dt =\frac{4t}{1+t^{2}}+4\arctan t.$$

Comments.

1. The Euler substitutions are more general, but when the integrand is a rational function of $x$ and $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, $\sqrt{x^2-a^2}$, with $a>0$, trigonometric and hyperbolic substitutions based on the trigonometric identities $$1-\sin^2 x=\cos^2 x,\quad 1+\tan^2 x=\sec^2x, \quad \sec^2x-1=\tan^2 x$$ and on the hyperbolic identities $$1-\tanh^2 x=\operatorname{sech}^2 x,\quad 1+\sinh^2 x=\cosh^2 x, \quad\cosh^2 x-1=\sinh^2 x$$ are in general faster. For example for $\sqrt{a^2-x^2}$ you can use $x=a\sin t$ or $x=a\tanh t$.
2. Note that if you use the trigonometric substitution $x=2\sin t-1$ suggested by
   lab bhattacharjee, instead of the one I indicate above, the two constants of integrations are not equal.

Answer 2

HINT:

Using Trigonometric substitution

as $\displaystyle-x^2-2x+3=4-(x+1)^2,$ set $x+1=2\sin\theta$