Question:
Integrate the function: \(\sqrt{x^2+4x+1}\)
Integrate the function: \(\sqrt{x^2+4x+1}\)
Updated On: Oct 4, 2023
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Solution and Explanation
Let \(I=\int \sqrt{x^2+4x+1}\,dx\)
=\(\int \sqrt{(x^2+4x+4)-3}dx\)
=\(\int \sqrt{(x+2)^2-(\sqrt3)^2}dx\)
It is known that,\(\int \sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log \mid x+\sqrt{x^2-a^2}\mid+C\)
∴=\(I=\frac{(x+2)}{2}\sqrt{x^2+4x+1-}\frac{3}{2}\log\mid (x+2)+\sqrt{x^2+4x+1}\mid+C\)
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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
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These are tabulated below along with the meaning of each part.
