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Question:
\(∫\frac{dx}{x^2+2x+2}\) equals
\(∫\frac{dx}{x^2+2x+2}\) equals
Updated On: Feb 29, 2024
\(x tan^{-1} (x+1)+ C\)
\(tan^{-1} (x+1)+ C\)
\((x+1) tan^{-1} x + C\)
\(tan^{-1} x + C\)
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The Correct Option is B
Solution and Explanation
The correct answer is \( [tan^{-1}(x+1)]+C\)
\(∫\frac{dx}{x^2+2x+2} = ∫\frac{dx}{(x^2+2x+1)}+1\)
\(= ∫\frac{1}{(x+1)^2+(1)^2} dx\)
\(= [tan^{-1}(x+1)]+C\)
Hence, the correct Answer is B
\(∫\frac{dx}{x^2+2x+2} = ∫\frac{dx}{(x^2+2x+1)}+1\)
\(= ∫\frac{1}{(x+1)^2+(1)^2} dx\)
\(= [tan^{-1}(x+1)]+C\)
Hence, the correct Answer is B
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