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 ?
 A: $$
\int\frac{x+1}{\sqrt{1-x^2}} \, dx = \int \frac x {\sqrt{1-x^2}} \, dx + \int \frac 1 {\sqrt{1-x^2}} \, dx
$$
Do the first integral by writing $u=1-x^2$ so that $x\,dx = -\frac 1 2\, du$.  Use a trigonometric substitution for the second one.
A: 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 $y = \sin^{-1}x$.
A: Alternative approach. Since $1-x^2=(1-x)(1+x)$, the integral can be written as:
$$ I=\int\sqrt{\frac{1+x}{1-x}}\,dx $$
and by setting $x=\frac{y-1}{y+1}$, then $y=z^2$, we get:
$$ I = \int \frac{2\sqrt{y}}{(1+y)^2}\,dy = 4\int \frac{z^2}{(1+z^2)^2}\,dz =C-\frac{2z}{1+z^2}+2\arctan(z).$$
A: $$\int  \frac { x+1 }{ \sqrt { 1-x^{ 2 } }  } dx=-\left( \int  \frac { -x }{ \sqrt { 1-x^{ 2 } }  } dx-\int { \frac { dx }{ \sqrt { 1-x^{ 2 } }  }  }  \right) =\\ =-\frac { 1 }{ 2 } \int  \frac { d\left( 1-{ x }^{ 2 } \right)  }{ \sqrt { 1-x^{ 2 } }  } +\int { \frac { dx }{ \sqrt { 1-x^{ 2 } }  } =-\sqrt { 1-x^{ 2 } } +\arcsin { x } +C } $$
A: $$\int\frac{x+1}{\sqrt{1-x^2}} dx = \int \frac x{\sqrt{1-x^2}}~dx + \int \frac 1{\sqrt{1-x^2}}~dx$$
The first integral can be done with a $u$-substitution $u = 1-x^2$ (so that $du = -2x~dx$). The second integral is $\arcsin x$ since the derivative of $\arcsin x$ is $\frac 1{\sqrt{1-x^2}}$.
A: This is to expand on Andre's comment. You can technically do this without explicitly using a trigonometric substitution if you define $\arcsin$ in a particular way. It turns out that one can use the definition $$\arcsin x := \int_0^x \frac{1}{\sqrt{1-x^2}}\, dx$$
as a starting point for the theory of the trigonometric functions. 
Of course, regardless of how you define $\arcsin$, you could note that its derivative is $\frac{1}{\sqrt{1-x^2}}$ in which case it's trivial. For most integrals, working backwards isn't effective, but honestly the derivative of $\arcsin$ should simply be memorized. 
A: The antiderivative is defined uniquely (up to an additive constant) and involves trigonometric functions.  Not sure how you can get to that without using trigonometric functions.
A: Using $\displaystyle
\int\frac{x+1}{\sqrt{1-x^2}} \, dx = \int \frac x {\sqrt{1-x^2}} \, dx + \int \frac 1 {\sqrt{1-x^2}} \, dx$, 
you could substitute $u=1-x^2, du=-2x dx$ in the first term 
and let $x=\tanh u, dx=\text{sech} ^2 u\, du$ in the second term to get
$\displaystyle -\frac{1}{2}\int\frac{1}{\sqrt{u}}du+\int\text{sech }u \,du=-\sqrt{1-x^2}+\int\frac{\cosh u}{1+\sinh^2 u}\,du$
$\displaystyle=-\sqrt{1-x^2}+\arctan\frac{x}{\sqrt{1-x^2}}+C$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\ul{Without}}$ using any trigonometric substitution !!! :

With the sub$\ds{\ldots\ x \equiv \half\pars{t - {1 \over t}}\ic\quad\imp\quad
t = \root{1 - x^{2}} - x\ic}$

\begin{align}
\int{x + 1 \over \root{1 -x^{2}}}\,\dd x & =
\int\pars{-\,\half + {1 \over 2t^{2}} + {\ic \over t}}\,\dd t =
-\,\half\,t - {1 \over 2t} + \ln\pars{t}\ic
\\[4mm] & =
-\,\half\,\root{1 - x^{2}} + \half\,x\ic - \half\,{1 \over \root{1 - x^{2}} - x\ic} +
\ln\pars{\root{1 - x^{2}} - x\ic}\ic
\\[4mm] & =
-\,\half\,\root{1 - x^{2}} + \half\,x\ic - \half\pars{\root{1 - x^{2}} + x\ic} +
\arctan\pars{x \over \root{1 - x^{2}}}
\\[4mm] & =
\color{#f00}{-\root{1 - x^{2}} + \arctan\pars{x \over \root{1 - x^{2}}}} +
\pars{~\mbox{a constant}~}
\end{align}
