Question

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Integrate the following function with respect to x

$\frac{1}{x( {x}^{5} - 1)}$
​

Answer 1

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

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1}{x( {x}^{5} - 1)} \: dx}$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{1}{x\bigg[ {x}^{5} - \dfrac{ {x}^{5} }{ {x}^{5} } \bigg]} \: dx$

can be further rewritten after taking common from denominator as

$\rm \: = \: \displaystyle\int\rm \frac{1}{x \times {x}^{5} \bigg[ 1 - \dfrac{ 1}{ {x}^{5} } \bigg]} \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{1}{{x}^{6} \bigg[ 1 - \dfrac{ 1}{ {x}^{5} } \bigg]} \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{ {x}^{ - 6} }{ 1 - \dfrac{ 1}{ {x}^{5} } } \: dx$

Now, we use method of Substitution to evaluate such integral.

So, Substitute

$\red{\rm :\longmapsto\:1 - \dfrac{1}{ {x}^{5} } = y}$

$\red{\rm :\longmapsto\:1 - {x}^{ - 5} = y}$

$\red{\rm :\longmapsto\: 5 {x}^{ - 5 - 1} \: dx = dy}$

$\red{\rm :\longmapsto\: 5 {x}^{ - 6} \: dx = dy}$

$\red{\rm :\longmapsto\: {x}^{ - 6} \: dx = \dfrac{dy}{ 5} }$

So, above integral can be rewritten as

$\rm \: = \:\displaystyle\int\rm \dfrac{dy}{ 5y}$

$\rm \: = \: \: \dfrac{1}{5} \:\displaystyle\int\rm \dfrac{dy}{y}$

$\rm \: = \: \: \dfrac{1}{5} log |y| + c$

$\rm \: = \: \: \dfrac{1}{5} log \bigg|1 - \dfrac{1}{ {x}^{5} } \bigg| + c$

$\rm \: = \: \: \dfrac{1}{5} log \bigg| \dfrac{ {x}^{5} - 1}{ {x}^{5} } \bigg| + c$

Hence,

$\boxed{ \tt{ \: \displaystyle\int\rm \: \frac{1}{x( {x}^{5} - 1)} \: dx \: = \: \dfrac{1}{5} log \bigg| \dfrac{ {x}^{5} - 1}{ {x}^{5}} \bigg| + c \: }}$

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Additional Information :-

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