Calculus: Indefinite Integrals of x^n / (x^2 + 1)
Task: Calculate the indefinite integrals taking the form of:
∫ x^n / (x^2 + 1) dx for n for n = 0, 1, 2, 3, 4, 5
∫ x^0 / (x^2 + 1) dx (n = 0)
∫ x^0 / (x^2 + 1) dx
= ∫ 1 / (x^2 + 1) dx
= arctan(x) + C
where C is a constant. This is a simple integral where we just have the derivative of arctan(x) in the integrand.
∫ x^1 / (x^2 + 1) dx (n = 1)
∫ x^1 / (x^2 + 1) dx
= ∫ x / (x^2 + 1) dx
Let u = x^2 + 1, then:
du = 2 * x dx
1 /2 du = x * dx.
Continuing:
= ∫ 1 / 2 * 1 / u du
= 1 / 2 * ∫ 1 / u du
= 1 / 2 * (ln |u| + C)
Substituting back:
= 1 / 2 * (ln |x^2 + 1| + C)
Since C is an arbitrary constant:
= 1 / 2 * ln |x^2 + 1| + C
∫ x^2 / (x^2 + 1) dx (n = 2)
∫ x^2 / (x^2 + 1) dx
It took me a bit to figure this out…. Start by adding and subtracting 1:
= ∫ (x^2 + 1 – 1) / (x^2 + 1) dx
= ∫ (x^2 + 1) / (x^2 + 1) - 1 / (x^2 + 1) dx
= ∫ 1 - 1 / (x^2 + 1) dx
= x – arctan(x) + C
∫ x^3 / (x^2 + 1) dx (n = 3)
∫ x^3 / (x^2 + 1) dx
Since the degree of the numerator is greater than the degree of the denominator, we can divide as so:
x^3 / (x^2 + 1) = x – x / (x^2 + 1)
This is going to be case from here on out.
∫ x^3 / (x^2 + 1) dx
= ∫ x – x / (x^2 + 1) dx
= x^2 / 2 – 1 / 2 * ln |x^2 + 1 | + C
∫ x^4 / (x^2 + 1) dx (n = 4)
x^4 / (x^2 + 1) = x^2 – 1 + 1 / (x^2 + 1)
∫ x^4 / (x^2 + 1) dx
= ∫ x^2 – 1 + 1 / (x^2 + 1) dx
= x^3 / 3 – x + arctan(x) + C
∫ x^5 / (x^2 + 1) dx (n = 5)
x^5 / (x^2 + 1) = x^3 – x + x / (x^2 + 1)
∫ x^5 / (x^2 + 1) dx
= ∫ x^3 – x + x / (x^2 + 1) dx
= x^4 / 4 – x^2 / 2 + 1 / 2 * ln |x^2 + 1| + C
Interesting point, I put this problem to Wolfram Alpha:
∫ x^5 / (x^2 + 1) dx
= 1 / 4 ((x^2 – 1)^2 + 2 * log(x^2 + 1)) + C
Note in computer language log(x) = ln (x), and x^2 + 1 > 0 for all x. (x^2 + 1) = |x^2 + 1|.
Multiply and simplify:
= 1 / 4 ((x^4 – 2 * x^2 + 1) + 2 * log(x^2 + 1)) + C
= x^4 / 4 – x^2 / 2 + 1 / 4 + 1 / 2 * ln |x^2 + 1| + C
Again, C is an arbitrary constant, hence C + 1 / 4 is also a constant.
= x^4 / 4 – x^2 / 2 + 1 / 2 * ln |x^2 + 1| + C’
(where C’ = C + 1 / 4)
So answers match. We’re OK. Sigh of relief.
∫ x^6 / (x^2 + 1) dx (n = 6)
x^6 / (x^2 + 1) = x^4 – x^2 + 1 – 1 / (x^2 + 1)
∫ x^6 / (x^2 + 1) dx
= ∫ x^4 – x^2 + 1 – 1 / (x^2 + 1) dx
= x^5 / 5 – x^3 / 3 + x – arctan(x) + C
Observations
I present this without proof. From what the pattern suggests:
∫ x^n / (x^2 + 1) dx
If n is odd, the integral ends in ± 1 / 2 * ln |x^2 + 1| + C
If n is even, the integral ends in ± arctan(x) + C
Hope you find this helpful. Welcome to March,
Eddie
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