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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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
