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The value of $\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}}$isa. $\dfrac{{\text{1}}}{{\text{n}}}{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{{{\text{x}}^{\text{n}}}{\text{ + 1}}}}{\text{) + C}}$b. ${\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}} + 1}}{{{{\text{x}}^{\text{n}}}}}{\text{) + C}}$c. $\dfrac{{\text{1}}}{{\text{n}}}{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{n}}}}}{\text{) + C}}$d. ${\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{{{\text{x}}^{\text{n}}} + 1}}{\text{) + C}}$

Last updated date: 13th Jun 2024
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Hint: First multiply numerator and denominator by ${{\text{x}}^{{\text{n - 1}}}}$and assume ${{\text{x}}^{\text{n}}}{\text{ = t}}$ and simplifying the problem. Then the integration will turn out to be a easier fraction to deal with.

Complete step by step solution: Given,
$\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}}$
On Multiplying denominator and numerator by ${{\text{x}}^{{\text{n - 1}}}}$,
$= \int {\dfrac{{{{\text{x}}^{{\text{n - 1}}}}{\text{dx}}}}{{{{\text{x}}^{\text{n}}}{\text{(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}}$
Now, let, ${{\text{x}}^{\text{n}}}{\text{ = t}}$
$\Rightarrow {{\text{x}}^{{\text{n - 1}}}}{\text{dx = }}\dfrac{{{\text{dt}}}}{{\text{n}}}$,
So we get,
${\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}}$………(1)

Now, we use the form of partial fractions to find the integral,
$\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{t + 1}} \\ \Rightarrow \dfrac{1}{{t(t + 1)}} = \dfrac{{A(t + 1) + Bt}}{{t(t + 1)}} \\$
So,
$A(t + 1) + Bt$= 1
If we take, $t = 0$ then,
$A(0 + 1) + B.0 = 1$ gives us $A = 1$
And also for , $t = - 1$
$A( - 1 + 1) + B. - 1 = 1$ gives us $B = - 1$
Now equation 1 can be written as,
$= \dfrac{1}{n}[\int {(\dfrac{1}{t} - \dfrac{1}{{t + 1}})dt}$
If we integrate now, we have, as integral of $\dfrac{1}{t}$ is $\log t$,
$\Rightarrow \dfrac{1}{n}[\log t - \log (t + 1)]$
Now, also,
`$\log a - \log b = \log \dfrac{a}{b}$
$\Rightarrow \dfrac{1}{n}[\log (\dfrac{t}{{t + 1}})]$
Putting the value of t we have,
$= \dfrac{1}{n}[\log (\dfrac{{{x^n}}}{{{x^n} + 1}})]$
So, we get, $\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}}$$= \dfrac{1}{n}[\log (\dfrac{{{x^n}}}{{{x^n} + 1}})]+C$

Note: $\dfrac{1}{{t(t + 1)}}$can be directly written as sum of two fractions $\dfrac{1}{t}, - \dfrac{1}{{t + 1}}$as $\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{t + 1}}$. We can also do this integration in another way by multiplying the denominator and numerator by ${x^n} - 1$.