Question

Evaluate the following

$\int \: \frac{dx}{ x\sqrt{ {x}^{n} - 1} }$
​

Answer 1

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

Given integral is

$\displaystyle \int \: \frac{1}{x \sqrt{ {x}^{n} - 1 } } \: dx$

To evaluate this integral, we use Method of Substitution

So, Substitute

$\rm \: \sqrt{ {x}^{n} - 1} = y$

On squaring both sides, we get

$\rm \: {x}^{n} - 1 = {y}^{2}$

$\rm \: {x}^{n} = {y}^{2} + 1$

So, on differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}{x}^{n} = \dfrac{d}{dx} {y}^{2} + \dfrac{d}{dx} 1$

$\rm \: {nx}^{n - 1} = 2y\dfrac{dy}{dx} + 0$

$\rm \: {nx}^{n - 1} = 2y\dfrac{dy}{dx}$

$\rm \: dx = \dfrac{2y}{{nx}^{n - 1} } \: dy$

can be further rewritten as

$\rm \: dx = \dfrac{2xy}{{nx}^{n} } \: dy$

$\rm \: \dfrac{dx}{x} = \dfrac{2y}{n( {y}^{2} + 1)} \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle \int \rm \frac{2y \: dy}{n \: ( {y}^{2} + 1) \: y }$

$\rm \: = \: \dfrac{2}{n} \displaystyle \int \rm \frac{\: dy}{\: {y}^{2} + 1\:}$

$\rm \: = \: \dfrac{2}{n} {tan}^{ - 1}y + c$

$\rm \: = \: \dfrac{2}{n} {tan}^{ - 1} \sqrt{ {x}^{n} - 1 } + c$

Hence,

$\rm\implies \boxed{\tt{ \:\displaystyle \int \rm \: \frac{dx}{x \sqrt{ {x}^{n} - 1} } = \: \dfrac{2}{n} {tan}^{ - 1} \sqrt{ {x}^{n} - 1 } + c}}$

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ADDITIONAL INFORMATION

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

Answer 2

Additional Information

f(x). int f(x)dx

1. k kx+c

2. sinx -cos x+c

3. cosx sin x+c

4. sec²x ten x+c

5. cosec²x -cot x+c

6. secxtanx sec x+c