Integral Calculus Question 194
Question: If $ \int{\frac{dx}{x(x^{n}+1)}=A\log | \frac{x^{n}+1}{x^{n}} |+B,B\in R} $ . Then
Options:
A) $ A=\frac{1}{2} $
B) $ A=-1 $
C) $ A=-\frac{1}{n} $
D) $ A=\frac{1}{2n} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int{\frac{dx}{x(x^{n}+1)}=\int{\frac{dx}{{x^{n+1}}( 1+\frac{1}{x^{n}} )}}} $ Put $ 1+\frac{1}{x^{n}}=t\Rightarrow -\frac{n}{{x^{n+1}}}dx=dt $ $ I=-\frac{1}{n}\int{\frac{dt}{t}=-\frac{1}{n}lnt+C=-\frac{1}{n}ln( 1+\frac{1}{x^{n}} )+C} $ $ I=-\frac{1}{n}ln( \frac{x^{n}+1}{x^{n}} )+C\therefore A=-\frac{1}{n} $
BETA
