What should I substitute here to solve the integral.

Question

So this is the sum I am banging my head against for weeks but cannot find the perfect substitution.

$ \int_0^{\frac 1 {\sqrt3}}\sqrt{x+\sqrt{x^2+1}}\,dx \;$ is equal to

(1) $\frac 13 \;\;\; \;\;\;\;\; (2) \frac23$

(3) $ \frac 12\int_1^{\sqrt3}(\sqrt x + \frac 1{\sqrt{x^3}})\,dx \;\;\;\;\;\;\;\;\;\;\; (4)\frac34\int_1^\sqrt3(\sqrt x + \frac 1{\sqrt{x^3}})\,dx $

I got some hints from the limits which are 0 and $\frac1{\sqrt3} $ which are values in $\tan x$ function. Substituting $x = \tan(\alpha) $ which comes out to be

$$\int_0^{\frac{\pi}6}\sqrt{\tan\alpha+ \sec\alpha} \,\sec^2\alpha\,d\alpha $$

I tried to do rationalization but still could not get any hint on what to substitute. This is a multiple-choice question by the way. I don't want the complete answer just some hints about what substitution can I use.

Answer 1

You could try $x = \sinh t$. (Then $1+x^2 = \cosh^2 t$.)

Answer 2

Let $y=\sqrt{x+\sqrt{x^2+1}}.$ Then, $(y^2-x)^2=x^2+1$ hence $$x=\frac{y^2-y^{-2}}2,\quad dx=(y+y^{-3})\,dy,$$ $$\begin{align}\int_0^{\frac1{\sqrt3}}\sqrt{x+\sqrt{x^2+1}}\,dx&=\int_1^{3^{1/4}}(y^2+y^{-2})\,dy\\ &=\left[\frac{y^3}3-\frac1y\right]_1^{3^{1/4}}\\&=\frac23. \end{align}$$

Answer 3

Substitute $u=\sqrt{x^2+1}+x$, then $$du=(\frac{x}{\sqrt{x^2+1}}+1)dx$$ So $$(u-x)^2=x^2+1\implies u^2-2ux=1\implies x=\frac{u^2-1}{2u}$$ $$du = \frac{2u^2}{u^2+1}dx$$ $$\int\sqrt{x+\sqrt{x^2+1}}\,dx =\int \sqrt u \cdot\frac{u^2+1}{2u^2}du=\frac12\int (u^{\frac12}+u^{-\frac32})du\\=\frac13u^{\frac32}-u^{-\frac12}+C$$

$$\int_0^{\frac 1 {\sqrt3}}\sqrt{x+\sqrt{x^2+1}}\,dx =\frac12\int_1^{\sqrt3} (u^{\frac12}+u^{-\frac32})du\\=\frac13u^{\frac32}-u^{-\frac12}\Big|_1^{\sqrt3}=\frac13 3^{\frac34}-3^{-\frac14}-\frac13+1=\frac23$$

Answer 4

Let $I$ be your integral. Note that $$ \frac{\mathrm{d}}{\mathrm{d}\alpha}\sqrt{\tan\alpha+\sec\alpha}=\frac12\sec\alpha\sqrt{\tan\alpha+\sec\alpha} $$ So integrate by parts ($\mathrm{d}u=\sqrt{\tan\alpha+\sec\alpha}\sec\alpha\,\mathrm{d}\alpha$, $v=\sec\alpha$) $$ I=[2\sqrt{\tan\alpha+\sec\alpha}\sec\alpha]_0^{\pi/6}-2\int_0^{\pi/6}\sqrt{\tan\alpha+\sec\alpha}\sec\alpha\tan\alpha\,\mathrm{d}\alpha $$ Integrate by parts again ($\mathrm{d}u=\sqrt{\tan\alpha+\sec\alpha}\sec\alpha\,\mathrm{d}\alpha$, $v=\tan\alpha$), and you get a $+4I$ in the RHS. So solving...