<s>[INST] Integrate $\int\frac{x+1}{\sqrt{1-x^2}} \; dx$ without using trigonometric substitution I want to solve: 

$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$

I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?
 [/INST] Given that the answer for $\int\frac1{\sqrt{1-x^2}}dx$ will be $\sin^{-1}x$ you can't hope to solve the integral without any clue about trig functions.  However, there is an easy way to deduce the answer just knowing that sine (and cosine) are the solutions to the harmonic differential equation
$$
\frac{d^2 y}{dx^2} = -y \implies y = A\sin(x+\delta)
$$
(We are actually going to use this insight with $x$ and $y$ swapped.)
Start from  $y= \int\frac1{\sqrt{1-x^2}}dx \implies \frac{dy}{dx}= \frac1{\sqrt{1-x^2}} \implies \frac{dx}{dy} = \sqrt{1-x^2}$
Now differentiate with respect to $y$ again, using the chain rule, with the knowledge that $\frac{dx}{dy} = \sqrt{1-x^2}$:
$$\frac{d}{dy} \sqrt{1-x^2} = \frac{-2x}{2\sqrt{1-x^2}} \frac{dx}{dy}=\frac{-2x}{2\sqrt{1-x^2}} \sqrt{1-x^2} = -x$$.
So if $y= \int\frac1{\sqrt{1-x^2}}dx $ then $x = A \sin(y+\delta)
$, or more properly, if $y= \int_0^x\frac1{\sqrt{1-t^2}}dt $ then $x = A \sin(y+\delta)$.
Now when $x=0$, the limits in the integral are identical so $\delta$ must be zero. (We know $A$ cannot be zero; that would make the integral identically zero.)  We are left with the probelm of determining $A$.
From the integrand, we can read off that at zero, $d\frac{dy}{dx} = \frac{dx}{dy} = 1$.  Since the slope of $\sin y$ at the origin is $1$, the multiplicative constant $A$ must be $1$.  
Thus if  $y= \int\frac1{\sqrt{1-x^2}}dx$ then $x=\sin y$ or $y = \sin^{-1}x$.
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