How to calculate $\int\frac{\sqrt {2x+1}}{\sqrt {x}} dx$

Question

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

I've tried substituting $t = \sqrt x$ and $x = t^2$ with $dx = 2t\ dt$ but I got stuck again at $2\int \sqrt{2t^2+1}$ .

Answer 1

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.

Answer 2

Substituting $$t=\sqrt{\frac{2x+1}{x}}$$ then we get $$x=\frac{1}{t^2-2}$$ and $$dx=-\frac{2t}{(t^2-2)^2}dt$$

Answer 3

$I =\displaystyle\int\frac{\sqrt {2x+1}}{\sqrt {x}} dx$

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 $C = \dfrac{C_1}{2}$

Answer 4

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 $a = \frac{1}{\sqrt{2}}$:

$$\int{\sqrt{x^2+a^2}}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\log|x+\sqrt{x^2+a^2}|}{2}+C$$

Answer 5

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.