find the integral of ln(x)/x^(6/5) from 10 to infinity?

Answer 1

[10, ∞]∫x^(-6/5) ln x dx

Let u=ln x, du=1/x dx, v=-5x^(-1/5), dv=x^(-6/5) dx. Then we have:

-5x^(-1/5) ln x |[10, ∞] + 5 [10, ∞]∫x^(-6/5) dx
-5x^(-1/5) ln x - 25 x^(-1/5) |[10, ∞]

Since polynomials dominate logarithms, we have that [x→∞]lim -5x^(-1/5) ln x - 25 x^(-1/5) = 0, so evaluating at the endpoints gives:

0 - (-5*10^(-1/5) ln 10 - 25 *10^(-1/5))
5*10^(-1/5) ln 10 + 25 *10^(-1/5)
10^(-1/5) * (5 ln 10 + 25) ≈ 23.039

Answer 2

ln(x)/(x)^(6/5)
Integration by parts
`l.(uv)dx = u`l.vdx - `l.{(du/dx) `l.vdx}dx
Let u =ln(x), and v=1/(x)^(6/5)
`l.ln(x)/x^(6/5) = ln(x).(-5)/x^(1/5) - `l.(1/x).{(-5)/x^(1/5)}dx
= -5ln(x)/x^(1/5) + 5`l.(1/x^6/5)dx
= -5ln(x)/x^(1/5) + 5.(-5)/x(1/5)
= -5ln(x)/x^(1/5) -25/x^(1/5)
Now put the ranges