Integral Calculus Question 101
Question: The value of $ \int{\frac{dx}{3-2x-x^{2}}} $ will be
[UPSEAT 1999]
Options:
A) $ \frac{1}{4}\log ,( \frac{3+x}{1-x} ) $
B) $ \frac{1}{3}\log ,( \frac{3+x}{1-x} ) $
C) $ \frac{1}{2}\log ,( \frac{3+x}{1-x} ) $
D) $ \log ,( \frac{1-x}{3+x} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int{\frac{dx}{3-2x-x^{2}}=\int{\frac{dx}{4-(x^{2}+2x+1)}}} $
$ =\int{\frac{dx}{4-{{(x+1)}^{2}}}} $ $ =\int{\frac{dt}{{{(2)}^{2}}-t^{2}}} $
Where $ x+1=t,,\therefore dx=dt $
$ \therefore I=\frac{1}{2.2}\log ( \frac{2+t}{2-t} ) $ $ =\frac{1}{4}\log ( \frac{2+x+1}{2-x-1} ) $ $ =\frac{1}{4}\log ( \frac{3+x}{1-x} ). $
BETA
