Integral-Calculus Question 509
Question: If $ \int{\frac{x^{2}-x+1}{x^{2}+1}{e^{{{\cot }^{-1}}x}}dx=A(x){e^{{{\cot }^{-1}}x}}+C} $ , then $ A(x) $ is equal to:
Options:
A) $ -x $
B) $ x $
C) $ \sqrt{1-x} $
D) $ \sqrt{1+x} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ I=\int{\frac{x^{2}-x+1}{x^{2}+1}.{e^{{{\cot }^{-1}}x}}dx} $ Put $ x=\cot t\Rightarrow -\cos ec^{2}tdt=dx $ Now, $ 1+{{\cot }^{2}}t=\cos ec^{2}t $
$ \therefore I=\int{\frac{e^{t}({{\cot }^{2}}t-\cot t+1)}{(1+{{\cot }^{2}}t)}}(-\cos ec^{2}t)dt $ $ =\int{e^{t}(\cot t-\cos ec^{2}t)dt=e^{t}\cot t+C} $ $ ={e^{{{\cot }^{-1}}x}}(x)+C\equiv A(x).{e^{{{\cot }^{-1}}x}}+C\Rightarrow A(x)=x $
BETA
