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FUNKER
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In an exam i stumbled when i saw this q
integrate:
1/(1+2x+x^2) dx
:grumpy:
help
integrate:
1/(1+2x+x^2) dx
:grumpy:
help
?himanshu121 said:[tex]\int \frac{dx}{1+2x+x^2}[/tex]
[tex]\int \frac{dx}{(1+x)^2} [/tex]
Whats the pro
The general formula for integrating 1/(1+2x+x^2) is ∫1/(a^2+bx+c) dx = 1/(b^2-4ac) ln |(ax+b)/(a√(a^2+bx+c))| + C, where a, b, and c are constants.
Yes, this integral can be solved using the substitution method. Let u = 1+2x+x^2, then du/dx = 2+2x and dx = du/(2+2x). Substituting these values into the integral, we get ∫1/u du = ln|u| + C = ln|1+2x+x^2| + C.
This integral is defined for all real values of x, except when the denominator (1+2x+x^2) equals 0. This occurs when x = -1 ± i, where i is the imaginary unit. Therefore, the range of values for which the integral is defined is (-∞, -1) U (-1, ∞).
This integral is convergent. As x approaches ±∞, the function 1/(1+2x+x^2) approaches 0, which means the area under the curve is finite.
Yes, this integral can also be solved using partial fractions decomposition. The function 1/(1+2x+x^2) can be written as 1/(x+1)^2, which can be decomposed into the sum of two fractions with denominators (x+1) and (x+1)^2. The integral can then be solved by integrating each of these fractions separately.