Final answer:
To find the integral from 0 to 2x of ln(t³ˣ⁺¹), you can use the substitution method. After simplification and integration by parts, the result is [(3x+1)/3] * [ln(t³ˣ⁺¹)(t³ˣ⁺¹)^(-1/3)] - (1/3) * (t³ˣ⁺¹)^(-1/3) + C, where C is the constant of integration.
Step-by-step explanation:
To find the integral from 0 to 2x of ln(t³ˣ⁺¹), we can use the substitution method. Let u = t³ˣ⁺¹, then du = (3x + 1)t^(3x)dt. We can rewrite the integral as ∫ ln(u)(1/3)(u^(-1/3))/(3x+1) du. Simplifying, we get (1/3)(3x+1)∫ ln(u)u^(-1/3) du. Next, we can integrate by parts, letting dv = ln(u)u^(-1/3) du and u = (3x+1)/3. Integrating, we get [(3x+1)/3] * [ln(u)u^(-1/3)] - ∫ [(3x+1)/3] * (1/u) * (-1/3)u^(-4/3) du. Simplifying further and evaluating the integral, we get [(3x+1)/3] * [ln(u)u^(-1/3)] - (1/3) * u^(-1/3). Finally, substituting back u = t³ˣ⁺¹, we get the result [(3x+1)/3] * [ln(t³ˣ⁺¹)(t³ˣ⁺¹)^(-1/3)] - (1/3) * (t³ˣ⁺¹)^(-1/3) + C, where C is the constant of integration.