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Question: The value of \[\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]is...

The value of ∫dxx(xn + 1)\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} is
a. 1nlog(xnxn + 1) + C\dfrac{{\text{1}}}{{\text{n}}}{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{{{\text{x}}^{\text{n}}}{\text{ + 1}}}}{\text{) + C}}
b. log(xn+1xn) + C{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}} + 1}}{{{{\text{x}}^{\text{n}}}}}{\text{) + C}}
c. 1nlog(xn + 1xn) + C\dfrac{{\text{1}}}{{\text{n}}}{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{n}}}}}{\text{) + C}}
d. log(xnxn+1) + C{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{{{\text{x}}^{\text{n}}} + 1}}{\text{) + C}}

Explanation

Solution

First multiply numerator and denominator by xn - 1{{\text{x}}^{{\text{n - 1}}}}and assume xn = t{{\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,
∫dxx(xn + 1)\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}}
On Multiplying denominator and numerator by xn - 1{{\text{x}}^{{\text{n - 1}}}},
=∫xn - 1dxxn(xn + 1)= \int {\dfrac{{{{\text{x}}^{{\text{n - 1}}}}{\text{dx}}}}{{{{\text{x}}^{\text{n}}}{\text{(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}}
Now, let, xn = t{{\text{x}}^{\text{n}}}{\text{ = t}}
⇒xn - 1dx = dtn\Rightarrow {{\text{x}}^{{\text{n - 1}}}}{\text{dx = }}\dfrac{{{\text{dt}}}}{{\text{n}}},
So we get,
 = 1n∫dtt(t + 1){\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,

1t(t+1)=At+Bt+1 ⇒1t(t+1)=A(t+1)+Btt(t+1)  \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)+BtA(t + 1) + Bt= 1
If we take, t=0t = 0 then,
A(0+1)+B.0=1A(0 + 1) + B.0 = 1 gives us A=1A = 1
And also for , t=−1t = - 1
A(−1+1)+B.−1=1A( - 1 + 1) + B. - 1 = 1 gives us B=−1B = - 1
Now equation 1 can be written as,
=1n[∫(1t−1t+1)dt= \dfrac{1}{n}[\int {(\dfrac{1}{t} - \dfrac{1}{{t + 1}})dt}
If we integrate now, we have, as integral of 1t\dfrac{1}{t} is log⁡t\log t,
⇒1n[log⁡t−log⁡(t+1)]\Rightarrow \dfrac{1}{n}[\log t - \log (t + 1)]
Now, also,
`log⁡a−log⁡b=log⁡ab\log a - \log b = \log \dfrac{a}{b}
⇒1n[log⁡(tt+1)]\Rightarrow \dfrac{1}{n}[\log (\dfrac{t}{{t + 1}})]
Putting the value of t we have,
=1n[log⁡(xnxn+1)]= \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: 1t(t+1)\dfrac{1}{{t(t + 1)}}can be directly written as sum of two fractions 1t,−1t+1\dfrac{1}{t}, - \dfrac{1}{{t + 1}}as 1t(t+1)=1t−1t+1\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 xn−1{x^n} - 1.