∫ (1)/(x²+1) dx
To find the integral of 1/(x^2 + 1) dx, we can make use of the arctan function
To find the integral of 1/(x^2 + 1) dx, we can make use of the arctan function.
Let’s begin by examining the denominator, x^2 + 1. This is a quadratic expression without any real roots. To simplify integration, we can rewrite the denominator as (x + i)(x – i), where i is the imaginary unit (√-1).
Next, we can perform a partial fraction decomposition. The expression 1/(x^2 + 1) can be rewritten as A/(x + i) + B/(x – i), where A and B are undetermined constants.
To determine the values of A and B, we can multiply the entire equation by x^2 + 1 and simplify:
1 = A(x – i) + B(x + i)
1 = Ax – Ai + Bx + Bi
1 = (A + B)x + (B – A)i
Equating the real and imaginary parts, we can form a system of equations:
(A + B) = 0 (Equation 1)
(B – A) = 1 (Equation 2)
Solving Equation 1 for B, we have B = -A. Substituting this value into Equation 2, we get -A – A = 2A = 1. Therefore, A = 1/2, and B = -1/2.
Now that we have our partial fraction decomposition, we can rewrite the integral as follows:
∫ (1)/(x^2 + 1) dx = ∫ (1/2)/(x + i) dx + ∫ (-1/2)/(x – i) dx
To integrate the first term, we can use the substitution method. Let u = x + i, then du = dx. Substituting back into the equation, we have:
∫ (1/2)/(x + i) dx = ∫ (1/2)/u du = (1/2) ln|u| + C1
Since u = x + i, we can simplify the natural logarithm:
= (1/2) ln|x + i| + C1
Now, integrating the second term follows a similar process. Let v = x – i, then dv = dx. Substituting back in:
∫ (-1/2)/(x – i) dx = ∫ (-1/2)/v dv = (-1/2) ln|v| + C2
Substituting v = x – i, we have:
= (-1/2) ln|x – i| + C2
Combining the two integrals, we get:
∫ (1)/(x^2 + 1) dx = (1/2) ln|x + i| – (1/2) ln|x – i| + C
Simplifying further, we can use the logarithmic identity ln(a) – ln(b) = ln(a/b):
= (1/2) ln|(x + i)/(x – i)| + C
Since (x + i)/(x – i) is a complex number, we can simplify this expression by multiplying the numerator and denominator by the conjugate of the denominator:
= (1/2) ln[(x + i)(x + i)]/[x^2 + (-i)^2] + C
= (1/2) ln[(x^2 + 1)/(x^2 + 1)] + C
= (1/2) ln(1) + C
= (1/2) * 0 + C
= C
Thus, the final result is:
∫ (1)/(x^2 + 1) dx = C
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