Integral Calculus Question 231
Question: $ \int{\frac{{x^{n-1}}}{x^{2n}+a^{2}}dx}= $
Options:
A) $ \frac{1}{na}{{\tan }^{-1}}( \frac{x^{n}}{a} )+C $
B) $ \frac{n}{a}{{\tan }^{-1}}( \frac{x^{n}}{a} )+C $
C) $ \frac{n}{a}{{\sin }^{-1}}( \frac{x^{n}}{a} )+C $
D) $ \frac{n}{a}{{\cos }^{-1}}( \frac{x^{n}}{a} )+C $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ I=\int{\frac{{x^{n-1}}dx}{x^{2n}+a^{2}}} $ Let $ x^{n}=t\Rightarrow n.{x^{n-1}}dx=dt $
$ \therefore I=\int{\frac{1}{n}.\frac{dt}{t^{2}+a^{2}}=\frac{1}{n}.\frac{1}{a}{{\tan }^{-1}}( \frac{t}{a} )+C} $ $ =\frac{1}{na}{{\tan }^{-1}}[ \frac{x^{n}}{a} ]+C $
BETA
