Question

$\displaystyle\int \sf\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \: dx$​

Answer 1

Answer:

$\huge \pink{\frac{ {x}^{5} }{5} + C}$

Step-by-step explanation:

Let

$f(x) = \sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \\ ....eq \: 1 \\ multiply \: both \: sides \: by \: x \\ \\ xf(x) = x\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \\ \\ = \sqrt{ {x}^{2} .{x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \\ \\ = \sqrt{ x.{x}^{3} \sqrt{ {x}^{2}. {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \\ \\ xf(x)= \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} \sqrt{x...} }}} \\ ....eq \: 2$

Square equation 1

$f {}^{2} (x) = { {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \\ \\ from \: eq \: 2 \\ \\ f {}^{2} (x) = {x}^{3} (xf(x)) \\ \\ \implies \: f {}^{2} (x) = {x}^{4} f(x) \\ \\ assuming \: f(x) \neq0 \\ \\ f(x) = {x}^{4}$

Integral

$\int \: f(x)dx = \int {x}^{4} dx = \frac{ {x}^{5} }{5} + C$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int \sf\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} ...} }}} \: dx$

To evaluate this integral, Let first evaluate

$\rm :\longmapsto\: \sf\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} ...} }}}$

Let assume that

$\rm :\longmapsto\:y = \sf\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} ...} }}}$

So, y should not be equal to zero.

can be rewritten on squaring as

$\rm :\longmapsto\: {y}^{2} = \sf\ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} ...} }}----(1)$

Now, Consider

$\rm :\longmapsto\:y = \sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}}$

So, can be rewritten as

$\rm :\longmapsto\:xy = x\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}}$

$\rm :\longmapsto\:xy = \sqrt{ {x}^{3}. {x}^{2} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}}$

$\rm :\longmapsto\:xy = \sqrt{ {x}^{4}.x \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}}$

$\rm :\longmapsto\:xy = \sqrt{ {x}^{4} \sqrt{ {x}^{4}. {x}^{2} \sqrt{ {x}^{5} \sqrt{x...} }}}$

$\rm :\longmapsto\:xy = \sqrt{ {x}^{4} \sqrt{ {x}^{5}. x \sqrt{ {x}^{5} \sqrt{x...} }}}$

$\rm :\longmapsto\:xy = \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{5}. {x}^{2} \sqrt{x...} }}}$

$\rm :\longmapsto\:xy = \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6}.x \sqrt{x...} }}}$

Continuing like this, we get

$\rm :\longmapsto\:xy = \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} \sqrt{x...} }}} - - (2)$

On substituting the value of (2) in (1), we get

$\rm :\longmapsto\: {y}^{2} = {x}^{3} \times xy$

$\rm :\longmapsto\: {y}^{2} = y {x}^{4}$

$\rm :\longmapsto\: y= {x}^{4}$

$\bf\implies \:\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} = {x}^{4}$

So,

$\rm :\longmapsto\:\displaystyle\int \sf\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{ {x}^{6} ...} }}} \: dx$

$\rm \:  =  \: \displaystyle\int {x}^{4} \: dx$

$\rm \:  =  \: \dfrac{ {x}^{4 + 1} }{4 + 1} + c$

$\rm \:  =  \: \dfrac{ {x}^{5} }{5} + c$

Hence,

$\red{\boxed{\tt{ \displaystyle\int \sf\sqrt{ {x}^{3} \sqrt{ {x}^{4} \sqrt{ {x}^{5} \sqrt{x...} }}} \: dx \: = \: \frac{ {x}^{5} }{5} + c \: }}}$

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Additional Information

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$