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) \[\frac{1}{2}\log ({{x}^{2}}+4x+5)-2{{\tan }^{-1}}(x+2)+c\]
Correct Answer: D
Solution :
\[I=\int{\frac{x\,\,dx}{{{x}^{2}}+4x+5}}\]\[=\int{\frac{x+2-2\,\,}{{{(x+2)}^{2}}+1}dx}\] \[=\frac{1}{2}\int{\frac{2(x+2)\,\,\,dx}{{{(x+2)}^{2}}+1}}-2\int{\frac{dx}{1+{{(x+2)}^{2}}}}\] \[=\frac{1}{2}\int{\frac{dt}{t}-2\int{\frac{dx}{1+{{(x+2)}^{2}}}}}\] [Put \[1+{{(x+2)}^{2}}=t\] in first expression Þ 2(x +2)dx = dt] \[=\frac{1}{2}\log t-2{{\tan }^{-1}}(x+2)+c\] \[=\frac{1}{2}\log ({{x}^{2}}+4x+5)-2{{\tan }^{-1}}(x+2)+c\].
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