Integral of x³÷2x⁴+1
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).
To find the integral of the given function,
∫(x³)/(2x⁴ + 1) dx,
we can start by simplifying the expression inside the integral. Notice that the denominator can be written as a sum of squares by factoring the denominator as a difference of squares:
2x⁴ + 1 = (√2x²)² + 1².
Using the formula for the difference of squares, we can rewrite this expression as:
2x⁴ + 1 = (√2x² + 1)(√2x² - 1).
Next, we can use the technique of partial fractions to break down the integrand:
(x³)/(2x⁴ + 1) = A/(√2x² + 1) + B/(√2x² - 1),
where A and B are constants to be determined.
To find the values of A and B, we can use a common denominator and equate numerators:
x³ = A(√2x² - 1) + B(√2x² + 1).
Expanding and rearranging the equation, we have:
x³ = (√2Ax² + B√2x²) - (A + B).
Matching coefficients of like powers of x on both sides, we get the following system of equations:
0x³: 0 = -A - B,
x²: 0 = √2A + √2B,
x¹: 1 = A.
From the first equation, we find that A = 1. Substituting this value into the second equation, we can solve for B:
0 = √2(1) + √2B,
-√2 = √2B,
B = -1.
Therefore, we have successfully determined the constants A = 1 and B = -1.
Now, we can rewrite the original integral using the partial fractions:
∫(x³)/(2x⁴ + 1) dx = ∫(1)/(√2x² + 1) dx - ∫(1)/(√2x² - 1) dx.
To evaluate these integrals, we use standard integral formulas:
∫(1)/(√a² + x²) dx = ln|x + √a² + √a² + x| / (2√a²),
where a is a non-negative constant.
Using this formula, we can integrate both terms:
∫(1)/(√2x² + 1) dx = ln|x + √2 + √2x| / (2√2),
∫(1)/(√2x² - 1) dx = ln|x + √2 - √2x| / (2√2).
Therefore, the final result is:
∫(x³)/(2x⁴ + 1) dx = ln|x + √2 + √2x| / (2√2) - ln|x + √2 - √2x| / (2√2) + C,
where C is the constant of integration.
To find the integral of the function f(x) = x³ / (2x⁴ + 1), you can use the substitution method. Let's proceed step-by-step:
Step 1: Substitute u = 2x⁴ + 1
Differentiate both sides with respect to x to find du:
du = 8x³ dx
Step 2: Rearrange the equation to solve for dx:
dx = du / (8x³)
Step 3: Substitute the values of u and dx into the integral:
∫ x³ / (2x⁴ + 1) dx = ∫ (x³ / u) * (du / 8x³)
Step 4: Simplify and cancel out the common terms:
∫ 1 / (8u) du
Step 5: Integrate the function:
∫ 1 / (8u) du = (1/8) * ∫ 1/u du
Step 6: Integrate the function further:
∫ 1/u du = (1/8) * ln|u| + C
Step 7: Substitute back the value of u:
(1/8) * ln|2x⁴ + 1| + C
Therefore, the integral of x³ / (2x⁴ + 1) is (1/8) * ln|2x⁴ + 1| + C, where C is the constant of integration.