Integral Calculus Question 133
Question: $ \int{\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}dx} $ is equal to
Options:
A) $ {{\tan }^{-1}}x+2{{\tan }^{-1}}( \frac{x}{2} )+C $
B) $ {{\tan }^{-1}}( \frac{x}{2} )-4{{\tan }^{-1}}x+C $
C) $ -\frac{1}{3}{{\tan }^{-1}}x+\frac{2}{3}{{\tan }^{-1}}( \frac{x}{2} )+C $
D) $ 4{{\tan }^{-1}}( \frac{x}{2} )-2{{\tan }^{-1}}x+C $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \int{\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}}dx $ . $ =\frac{1}{3}\int{[ \frac{4}{x^{2}+4}-\frac{1}{x^{2}+1} ]dx} $ $ =-\frac{1}{3}\int{\frac{1}{x^{2}+1}}dx+\frac{4}{3}\int{\frac{1}{x^{2}+4}dx} $ $ =-\frac{1}{3}{{\tan }^{-1}}x+\frac{4}{3}\times \frac{1}{2}{{\tan }^{-1}}( \frac{x}{2} )+C $ $ =-\frac{1}{3}{{\tan }^{-1}}x+\frac{2}{3}{{\tan }^{-1}}( \frac{x}{2} )+C $
BETA
