Question
Integral of x³÷2x⁴+1
Answers
GPT 3.5
To find the integral of x³ / (2x⁴+1), we can start by expressing the denominator as a power of x. Notice that 2x⁴ + 1 is the same as (x²)² + 1. Letting u = (x²), we can rewrite the integral as:
∫ (x³ / (2x⁴+1)) dx = ∫ (u² / (2u² + 1)) dx.
Now, let's take the derivative of u with respect to x. We have:
du/dx = d/dx (x²) = 2x.
Rearranging this equation, we get dx = (1/2x) du. Substituting this back into the integral, we have:
∫ (u² / (2u² + 1)) (1/2x) du = (1/2) ∫ (u² / (u² + 1/2)) (1/x) du.
Notice that we can further simplify the expression in the parentheses by factoring out 1/2 from the denominator:
∫ (u² / (u² + 1/2)) (1/x) du = (1/2) ∫ (u² / (1/2)(2u² + 1)) (1/x) du.
Now, the expression in the parentheses becomes 2 ∫ (u² / (2u² + 1)) (1/x) du. Let's rewrite this integral in terms of u:
∫ (u² / (2u² + 1)) (1/x) du = (1/2) ∫ (u² / (2u² + 1)) (1/x) du = (1/2) ∫ (u² / (2u² + 1)) (1/u²) du.
Now, notice that we have (u² / u²) in the integral, which simplifies to just 1:
∫ (u² / (2u² + 1)) (1/u²) du = (1/2) ∫ (1 / (2u² + 1)) du.
This integral can be evaluated as:
(1/2) ∫ (1 / (2u² + 1)) du = (1/2) arctan(sqrt(2u²)) + C,
where arctan(sqrt(2u²)) + C represents the antiderivative of 1 / (2u² + 1) with respect to u. Substituting back for u = (x²), we have:
(1/2) arctan(sqrt(2u²)) + C = (1/2) arctan(sqrt(2(x²))) + C,
which is the integral of x³ / (2x⁴ + 1).
∫ (x³ / (2x⁴+1)) dx = ∫ (u² / (2u² + 1)) dx.
Now, let's take the derivative of u with respect to x. We have:
du/dx = d/dx (x²) = 2x.
Rearranging this equation, we get dx = (1/2x) du. Substituting this back into the integral, we have:
∫ (u² / (2u² + 1)) (1/2x) du = (1/2) ∫ (u² / (u² + 1/2)) (1/x) du.
Notice that we can further simplify the expression in the parentheses by factoring out 1/2 from the denominator:
∫ (u² / (u² + 1/2)) (1/x) du = (1/2) ∫ (u² / (1/2)(2u² + 1)) (1/x) du.
Now, the expression in the parentheses becomes 2 ∫ (u² / (2u² + 1)) (1/x) du. Let's rewrite this integral in terms of u:
∫ (u² / (2u² + 1)) (1/x) du = (1/2) ∫ (u² / (2u² + 1)) (1/x) du = (1/2) ∫ (u² / (2u² + 1)) (1/u²) du.
Now, notice that we have (u² / u²) in the integral, which simplifies to just 1:
∫ (u² / (2u² + 1)) (1/u²) du = (1/2) ∫ (1 / (2u² + 1)) du.
This integral can be evaluated as:
(1/2) ∫ (1 / (2u² + 1)) du = (1/2) arctan(sqrt(2u²)) + C,
where arctan(sqrt(2u²)) + C represents the antiderivative of 1 / (2u² + 1) with respect to u. Substituting back for u = (x²), we have:
(1/2) arctan(sqrt(2u²)) + C = (1/2) arctan(sqrt(2(x²))) + C,
which is the integral of x³ / (2x⁴ + 1).