s- integral
s (p^5) lnp dp ?
= (lnp)[(p^6)/(6)] - s (i'm confused here)
Let ln p = u and p^5 dv = dv
du = 1/p and v = p^6/6
The integral can be written uv - v du =
(ln p*p^6)/6 - S (p^5/6)
= (1/6)ln p*p^6 - ???
Complete the last step
To complete the last step, you need to evaluate the remaining integral that you have. Recall that the integral of (p^5)/6 with respect to p is:
∫ (p^5)/6 dp
To integrate this, you can use the power rule for integration. The power rule states that if you have an expression of the form x^n, where n is any real number except -1, then the integral of x^n with respect to x is given by:
∫ x^n dx = (x^(n+1))/(n+1)
Applying this rule to our integral, we have:
∫ (p^5)/6 dp = (1/6) ∫ p^5 dp
Using the power rule, the integral becomes:
(1/6) * (p^(5+1))/(5+1) = (1/6) * (p^6)/6
Now, substituting back the original variable ln p, we have:
(1/6) * (ln p^6)/6 = (ln p^6)/36
Therefore, the final result of the integral is:
(ln p^6)/36