Final answer:
To evaluate the indefinite integral ∫(x²+1) ln(x³+3 x+1)/(x³+3 x+1) d x, we can use the method of integration by parts. First, we define u = ln(x³+3 x+1) and dv = (x²+1)/(x³+3 x+1) d x. After finding du and v, we apply the formula for integration by parts.
Step-by-step explanation:
To evaluate the indefinite integral ∫(x²+1) ln(x³+3 x+1)/(x³+3 x+1) d x, we can use the method of integration by parts. Let's define u = ln(x³+3 x+1) and dv = (x²+1)/(x³+3 x+1) d x.
Next, we need to find du and v:
- du = (1/(x³+3 x+1)) * (3 x² + 3)
- dv = (x²+1)/(x³+3 x+1) d x
- v = ∫(x²+1)/(x³+3 x+1) d x
Now we can apply the formula for integration by parts:
∫u dv = uv - ∫v du
Plugging in the values, we have:
∫(x²+1) ln(x³+3 x+1)/(x³+3 x+1) d x = ln(x³+3 x+1) * ∫(x²+1)/(x³+3 x+1) d x - ∫[(3 x² + 3)/(x³+3 x+1)] ln(x³+3 x+1) d x
The second term on the right-hand side is a bit complex, but the first term is something we can evaluate:
∫(x²+1)/(x³+3 x+1) d x = ∫(1/x + 2/(x²(x+3))) d x = ln(x) + ln(x²+3)
So the final result is:
∫(x²+1) ln(x³+3 x+1)/(x³+3 x+1) d x = ln(x³+3 x+1) * [ln(x) + ln(x²+3)] - ∫[(3 x² + 3)/(x³+3 x+1)] ln(x³+3 x+1) d x