Question

#Quality Question
@Integrals


Find the value of

$\huge \int \frac{dx}{x( {x}^{n} + 1)}$
​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{dx}{x( {x}^{n} + 1)}$

$= \int \frac{( {x}^{n} + 1 - {x}^{n}) dx}{x( {x}^{n} + 1)}$

$= \int \frac{( {x}^{n} + 1) dx}{x( {x}^{n} + 1)} - \int \frac{ {x}^{n} dx}{x( {x}^{n} + 1) }$

$= \int \frac{ dx}{x} - \int \frac{ {x}^{n - 1} dx}{( {x}^{n} + 1) }$

$= ln |x| - ln | {x}^{n} + 1| + C$

$= ln | \frac{x}{ {x}^{n} + 1 }| + C$

Answer 2

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

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{dx}{x( {x}^{n} + 1)}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{dx}{x. {x }^{n}(1 + {x}^{ - n}) }$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{dx}{ {x }^{n + 1}(1 + {x}^{ - n}) }$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{ {x}^{ - (n + 1)} }{1 + {x}^{ - n} }dx$

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

$\red{\rm :\longmapsto\:Put \:1 + {x}^{ - n} = y}$

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

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

On substituting all these values, we get

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

We know,

$\boxed{ \sf{ \:\displaystyle\int\rm \frac{1}{x}dx = logx + c}}$

$\rm \:  =  \:  \: - \: \dfrac{1}{n} \: log(y) + c$

$\rm \:  =  \:  \: - \: \dfrac{1}{n} \: log(1 + {x}^{ - n} ) + c$

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

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

Alternative Method

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{dx}{x( {x}^{n} + 1)}$

$\purple{\bf :\longmapsto\:Multiply \: and \: divide \: by \: {nx}^{n - 1}}$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{n {x}^{n - 1} }{{nx }^{n - 1}.x.({x}^{n} + 1) } dx$

$\rm \:  =  \:  \:\displaystyle\int\rm \dfrac{n {x}^{n - 1} }{{nx }^{n}.({x}^{n} + 1) } dx$

$\red{\rm :\longmapsto\:Put \:{x}^{n} = y}$

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

On substituting the values, we get

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

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

$\rm \:  =  \:  \: \dfrac{1}{n}\displaystyle\int\rm \dfrac{1 + y - y}{y(y + 1)}dy$

$\rm \:  =  \:  \: \dfrac{1}{n}\displaystyle\int\rm \dfrac{1 + y }{y(y + 1)}dy - \dfrac{1}{n}\displaystyle\int\rm \dfrac{y}{y(y + 1)}dy$

$\rm \:  =  \:  \: \dfrac{1}{n}\displaystyle\int\rm \dfrac{1}{y}dy - \dfrac{1}{n}\displaystyle\int\rm \dfrac{1}{y + 1}dy$

$\rm \:  =  \:  \: \dfrac{1}{n} log(y) - \dfrac{1}{n} log(y + 1) + c$

$\rm \:  =  \:  \: \dfrac{1}{n} log( {x}^{n} ) - \dfrac{1}{n} log( {x}^{n} + 1) + c$

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

Properties used :-

$\rm :\longmapsto\: log( {x}^{y} ) = y \: logx$

$\rm :\longmapsto\:log\bigg |\dfrac{y}{x} \bigg| = logy - logx$

$\rm :\longmapsto\: {x}^{m} \times {x}^{n} = {x}^{m + n}$

$\rm :\longmapsto\: {x}^{ - n} = \dfrac{1}{ {x}^{n} }$