Q.Integrate the following functions with respect to $x$:

Rationalise the integrand by multiplying by $\sqrt{1-x}/\sqrt{1-x}$, then split and integrate term by term.

Multiply numerator and denominator inside the square root by $(1-x)$:

$\sqrt{\dfrac{1-x}{1+x}}=\sqrt{\dfrac{(1-x)^2}{(1+x)(1-x)}}=\dfrac{1-x}{\sqrt{1-x^2}}=\dfrac{1}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}.$

Now integrate each piece:

$\int\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x,\qquad \int\dfrac{-x\,dx}{\sqrt{1-x^2}}=\sqrt{1-x^2}\ \ \left(\text{since }\dfrac{d}{dx}\sqrt{1-x^2}=\dfrac{-x}{\sqrt{1-x^2}}\right).$

Adding: $\displaystyle\int\sqrt{\dfrac{1-x}{1+x}}\,dx=\sin^{-1}x+\sqrt{1-x^2}+c$.

Writing $\cos 2x = \cos^2x(1-\tan^2x)$ converts the integrand into $\sec^2x/\sqrt{1-\tan^2x}$, and substituting $t=\tan x$ gives a standard $\int dt/\sqrt{1-t^2}$ integral.

$\dfrac{\sec x}{\sqrt{\cos 2x}} = \dfrac{1}{\cos x\sqrt{\cos 2x}}$.

Use $\cos 2x = \cos^2x - \sin^2x = \cos^2x(1-\tan^2x)$, so $\sqrt{\cos 2x} = \cos x\sqrt{1-\tan^2x}$ (taking $\cos x>0$).

Then $\dfrac{1}{\cos x\cdot\cos x\sqrt{1-\tan^2x}} = \dfrac{\sec^2x}{\sqrt{1-\tan^2x}}$.

Let $t=\tan x$, so $dt=\sec^2x\,dx$. The integral becomes $\displaystyle\int \dfrac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t)+c = \sin^{-1}(\tan x)+c$.

The substitution $u=e^x$ reduces the integral to $\int\frac{du}{u^2+1} = \tan^{-1}u+c$, giving $\tan^{-1}(e^x)+c$.

Let $u = e^x$, so $du = e^x\,dx$.

$\displaystyle\int \frac{e^x}{e^{2x}+1}\,dx = \int \frac{e^x\,dx}{(e^x)^2+1} = \int \frac{du}{u^2+1} = \tan^{-1}(u)+c = \tan^{-1}(e^x)+c$