Integral Calculus Question 74
Question: $ \int{\frac{xdx}{x^{2}+4x+5}=} $
[RPET 2002]
Options:
A) $ \frac{1}{2}\log (x^{2}+4x+5)+2{{\tan }^{-1}}(x)+c $
B) $ \frac{1}{2}\log (x^{2}+4x+5)-{{\tan }^{-1}}(x+2)+c $
C) $ \frac{1}{2}\log (x^{2}+4x+5)+{{\tan }^{-1}}(x+2)+c $
D) $ \frac{1}{2}\log (x^{2}+4x+5)-2{{\tan }^{-1}}(x+2)+c $
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Answer:
Correct Answer: D
Solution:
$ I=\int{\frac{xdx}{x^{2}+4x+5}} $ $ =\int{\frac{x+2-2}{{{(x+2)}^{2}}+1}dx} $ $ =\frac{1}{2}\int{\frac{2(x+2),dx}{{{(x+2)}^{2}}+1}}-2\int{\frac{dx}{1+{{(x+2)}^{2}}}} $ $ =\frac{1}{2}\int{\frac{dt}{t}-2\int{\frac{dx}{1+{{(x+2)}^{2}}}}} $ [Put $ 1+{{(x+2)}^{2}}=t $ in first expression Þ 2(x +2)dx = dt] $ =\frac{1}{2}\log t-2{{\tan }^{-1}}(x+2)+c $ $ =\frac{1}{2}\log (x^{2}+4x+5)-2{{\tan }^{-1}}(x+2)+c $ .
BETA
