Question

Solve the integral of :-
$\displaystyle\int \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{... } } } } \: \: d$
​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given integral.

$\displaystyle\rm= \int \sqrt{x + \sqrt{x + \sqrt{x + ... \infty } } } \: dx$

Let us assume that:

$\rm: \longmapsto y=\sqrt{x+ \sqrt{x + \sqrt{x + ... \infty } } }$

We can also write it as:

$\rm: \longmapsto y=\sqrt{x+y }$

Squaring both sides, we get:

$\rm: \longmapsto {y}^{2} =x+y$

$\rm: \longmapsto {y}^{2} - y - x = 0$

Using quadratic formula, we get:

$\rm: \longmapsto y = \dfrac{1 \pm \sqrt{ {( - 1)}^{2} - 4(1)( - x)} }{2}$

$\rm: \longmapsto y = \dfrac{1 \pm \sqrt{1 + 4x} }{2}$

But y cannot be negative. Therefore:

$\rm: \longmapsto y = \dfrac{1 + \sqrt{1 + 4x} }{2}$

Now, we got the value of y. Lets integrate.

$\displaystyle\rm= \int y\: dx$

$\displaystyle\rm= \int \dfrac{1 + \sqrt{4x + 1} }{2} \: dx$

$\displaystyle\rm= \dfrac{1}{2} \int 1 + \sqrt{4x + 1} \: dx$

$\displaystyle\rm= \dfrac{1}{2} \bigg[ \int 1 \: dx + \int \sqrt{4x + 1} \: dx \bigg]$

$\displaystyle\rm= \dfrac{1}{2} \bigg[ \int {x}^{0} \: dx + \int \sqrt{4x + 1} \: dx \bigg]$

As we know that:

$\displaystyle \rm : \longmapsto \int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} ,n \neq - 1$

We get:

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \int \sqrt{4x + 1} \: dx \bigg]$

Let us assume that:

$\rm : \longmapsto u = 4x + 1$

$\rm : \longmapsto du = 4 \: dx$

$\rm : \longmapsto dx = \dfrac{1}{4}du$

Therefore, the integral becomes:

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \int \dfrac{1}{4} \sqrt{u} \: du \bigg]$

We know that:

$\displaystyle \rm : \longmapsto \int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} ,n \neq - 1$

We get:

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \dfrac{1}{4} \dfrac{ {u}^{^{3}/_{2}} }{ ^{3}/_{2}} \bigg]$

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \dfrac{1}{4} \dfrac{2 {u}^{^{3}/_{2}} }{3} \bigg]$

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \dfrac{1}{2} \dfrac{{u}^{^{3}/_{2}} }{3} \bigg]$

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \dfrac{{u}^{^{3}/_{2}} }{6} \bigg]$

Substitute back u = 4x + 1, we get:

$\displaystyle\rm= \dfrac{1}{2} \bigg[x+ \dfrac{{(4x + 1)}^{^{3}/_{2}} }{6} \bigg]$

$\displaystyle\rm= \dfrac{1}{2} \times \dfrac{6x + (4x + 1) \sqrt{4x + 1} }{6}$

$\displaystyle\rm= \dfrac{6x + (4x + 1) \sqrt{4x + 1} }{12}$

Add the constant of integration:

$\displaystyle\rm= \dfrac{6x + (4x + 1) \sqrt{4x + 1} }{12} + C$

Therefore:

$\displaystyle\rm: \longmapsto\int \sqrt{x + \sqrt{x + \sqrt{x + ... \infty } } } \: dx \rm= \dfrac{6x + (4x + 1) \sqrt{4x + 1} }{12} + C$

Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf kx+C\\ \\ \sf sin(x)&\sf-cos(x)+C\\ \\ \sf cos(x)&\sf sin(x)+C\\ \\ \sf{sec}^{2}(x)&\sf tan(x)+C\\ \\ \sf{cosec}^{2}(x)&\sf-cot(x)+C\\ \\ \sf sec(x)\ tan(x)&\sf sec(x)+C\\ \\ \sf cosec(x)\ cot(x)&\sf-cosec(x)+C\\ \\ \sf tan(x)&\sf log(sec(x))+C\\ \\ \sf\dfrac{1}{x}&\sf log(x)+C\\ \\ \sf{e}^{x}&\sf{e}^{x}+C\\ \\ \sf x^{n},n\neq-1&\sf\dfrac{x^{n+1}}{n+1}+C\end{array}}$