How to calculate $\int\frac{\sqrt {2x+1}}{\sqrt {x}} dx$ $\int\frac{\sqrt {2x+1}}{\sqrt {x}} dx$
I've tried substituting $t = \sqrt x$ and $x = t^2$ with $dx = 2t\ dt$ but I got stuck again at $2\int \sqrt{2t^2+1}$ .
 A: Substituting $$t=\sqrt{\frac{2x+1}{x}}$$ then we get
$$x=\frac{1}{t^2-2}$$ and $$dx=-\frac{2t}{(t^2-2)^2}dt$$
A: While others have already answered the question, here is a take on using the same substitution that the OP was trying.
Substitute $t = \sqrt{x} \implies x = t^2$, and $\mathrm dx = 2\sqrt x\mathrm dt$.
$$\implies\int\dfrac{\sqrt{2x + 1}}{\sqrt x}\,\mathrm dx = 2\int\sqrt{2t^2 + 1}\,\mathrm dt$$
Substitute $t = \dfrac{\tan(s)}{\sqrt{2}}\implies s = \arctan{\sqrt2t}$, and $\mathrm dt = \dfrac{\sec^2s}{\sqrt2}\mathrm ds$. Also, $1 + \tan^2s = \sec^2s$.
$$\implies2\int\sqrt{2t^2 + 1}\,\mathrm dt = \dfrac2{\sqrt2}\int\sec^2s\sqrt{1 + \tan^2s}\,\mathrm ds = \sqrt2\int\sec^3s\,\mathrm ds$$
This integral is already solved by The Integrator.
A: $I =\displaystyle\int\frac{\sqrt {2x+1}}{\sqrt {x}} dx$
let $2x = \tan^2(z)\implies 2 dx = 2\tan(z)\sec^2(z)\,dz$
$I = \displaystyle\int\dfrac{\sqrt2\cdot\sec(z)}{\tan(z)}\tan(z)\sec^2(z)\,dz$
$I = \displaystyle\sqrt2\int\sec^3(z)\,dz$
$I =\sqrt2\displaystyle \bigg[\frac12\sec(z)\tan(z) +\frac12 \ln|\sec(z)+\tan(z)|\bigg]+C$
$I = \displaystyle\frac1{\sqrt2}\sqrt{2x}\cdot \sqrt{1+2x}+\frac1{\sqrt2}\ln|\sqrt{1+2x}+\sqrt{2x}|+C$
NOTE:
integral of $\sec^3(z)$;
$J = \displaystyle\int\sec^3(z)\,dz$
$J = \displaystyle \sec(z)\tan(z)-\int\sec(z)\tan^22(z)\,dz$
$J = \displaystyle \sec(z)\tan(z)-\int\sec(z)(\sec^2(z)-1)\,dz$
$J = \displaystyle\sec(z)\tan(z)- \int\sec^3(z)\,dz+\int\sec(z)\,dz$
$2J =\sec(z)\tan(z)+\ln|\sec(z)+\tan(z)|+C_1$
$J  = \displaystyle\frac12\sec(z)+\frac12\ln|\sec(z)+\tan(z)|+C$  $\qquad\qquad$ where $C = \dfrac{C_1}{2}$
A: Hint:
Once you arrive at $$2\int \sqrt{2t^2+1}\,dt = 2\sqrt{2}\int \sqrt{t^2+\frac12}\,dt$$ you can proceed with this general formula with $a = \frac{1}{\sqrt{2}}$:
$$\int{\sqrt{x^2+a^2}}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\log|x+\sqrt{x^2+a^2}|}{2}+C$$
A: Here's another approach using a sequence of simple substitutions. Certainly they could be combined to reduce the amount of work, but maybe it is helpful to see how one can proceed one small step at a time.
$$I = \int\frac{\sqrt {2x+1}}{\sqrt {x}}\ dx$$
First, simplify the root:
$$I = \int\sqrt{2+\frac1x}\ dx$$
Let's try $u=\frac1x$, so $x=\frac1u$ and $dx = -\frac1{u^2}\ du$:
$$I = -\int\frac{\sqrt{2+u}}{u^2}\ du$$
Now we try to simplify the square root further: let $v=2+u$, so $u=v-2$ and $du = dv$. Then we have:
$$I = -\int\frac{\sqrt v}{(v-2)^2}\ dv$$
To get rid of the root entirely now, we use $v=w^2, dv=2w\ dw$:
$$I = -2\int\frac{w^2}{(w^2-2)^2}\ dw$$
We are now effectively at the same state as in @DrSonnhardGraubner's answer. From here, use partial fractions and integrate.
