Question

Evaluate the following integral : \[\int{\dfrac{1}{x({{x}^{n}}+1)}}dx\]

Answer 1

Hint: Before solving integration you should know every formula of integration otherwise it will be difficult to solve the problems of integration. We can use the substitution method in integration. After that, we will convert this integration into a partial fraction and then we will solve the partial fraction using a suitable identity.

Complete step-by-step solution:
We have to integrate \[\int{\dfrac{1}{x({{x}^{n}}+1)}}dx\]………eq(1)
Now to solve this question we have to multiply eq(1) by \[{{x}^{n-1}}\] in both numerator and denominator. Which is as follows.
\[\int{\dfrac{{{x}^{n-1}}}{x({{x}^{n}}+1){{x}^{n-1}}}dx}\]
Now after multiplying by \[{{x}^{n-1}}\] in both numerator and denominator, we get the following results
\[I=\int{\dfrac{{{x}^{n-1}}}{({{x}^{n}}+1){{x}^{n}}}dx}\]………eq(2)
Substitution method should be done to solve this integration
Now we will put the value of \[{{x}^{n}}=t\] and will then differentiate this and then the following result is obtained
On differentiating both the sides
\[d({{x}^{n}})=dt\]
We get
\[\begin{align}
  & n{{x}^{n-1}}dx=dt \\
 & \Rightarrow {{x}^{n-1}}dx=\dfrac{dt}{n} \\
\end{align}\]……….eq(3)
 After doing substitution the integration is changed on putting eq(3) in eq(2), we get
\[I=\dfrac{1}{n}\int{\dfrac{dt}{t(t+1)}}\]
Now we will solve this with the help of a partial fraction and after applying partial fraction, the following result is obtained
Using the suitable identity of integration, \[\int{\dfrac{dx}{x(x+1)}=\dfrac{A}{x}+\dfrac{B}{x+1}}\]
Now we will use this identity in our integration, we get
\[\dfrac{1}{n}\int{\dfrac{dt}{t(t+1)}=\dfrac{A}{t}+\dfrac{B}{t+1}}\]……..eq(4)
\[\Rightarrow \dfrac{1}{t(t+1)}=\dfrac{A}{t}+\dfrac{B}{t+1}\]
Now we will take the LCM and will solve our question
\[\Rightarrow \dfrac{1}{t(t+1)}=\dfrac{A(t+1)+Bt}{t(t+1)}\]
On canceling out the denominator with each other so that there will be no fraction part and it will become easy for us to solve the integration.
We get,
\[\begin{align}
  & 1=A(t+1)+Bt \\
 & \Rightarrow 1=A+At+Bt \\
\end{align}\]
On comparing the above results, we will find the values of constants A and B to further solve our question
\[A=1\]
And, \[A+B=0\]
On putting the value of A in the above equation, we get the value of B as \[B=-1\]
On putting the values of both A and B in eq(4), we get
\[\dfrac{1}{n}\int{\dfrac{dt}{t(t+1)}=\dfrac{1}{n}\int{\left( \dfrac{1}{t}-\dfrac{1}{(t+1)} \right)}}dt\]
Now we will solve them separately as follows
\[\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}dt=\dfrac{1}{n}\left[ \int{\dfrac{1}{t}dt+\int{\dfrac{1}{t+1}dt}} \right]}\]
\[\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}dt=\dfrac{1}{n}\left[ \log t-\log \left| t+1 \right| \right]}\]
Using the property of logarithm
\[\log A-\log B=\log \left| \dfrac{A}{B} \right|\], we get the following results
\[\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}dt=\dfrac{1}{n}\log \left| \dfrac{t}{t+1} \right|}\]
Previously we have put on the value of \[{{x}^{n}}=t\], so now we will replace this value again to obtain the correct result.
\[\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}dt=\dfrac{1}{n}\log \left| \dfrac{{{x}^{n}}}{{{x}^{n}}+1} \right|}+C\]
Where C is any constant

Note: If the degree of the numerator is greater than the degree of the denominator or if the degree is the same then the fraction is known as an improper fraction and if the degree of the numerator is less than the degree of the denominator then it is known as a proper fraction. If we have linear equations in numerator and denominator then it can be solved by a partial fraction.