Answer:
[tex]f(u) = u[/tex]
Step-by-step explanation:
Given
[tex]\int {\ln(u)} \, du = u[\ln(u) - f(u)] + c[/tex]
Required
Find f(u)
To do this, we start by integrating the left-hand side
[tex]\int {\ln(u)} \, du = u\ln(u) - f(u)+ c[/tex]
Using integration by parts, we have:
[tex]\int {fg'} = fg - \int f'g[/tex]
So, we have:
[tex]f = \ln(u)[/tex]
Differentiate
[tex]f'=\frac{1}{u}[/tex]
[tex]g' = 1[/tex]
Integrate
[tex]g=u[/tex]
So:
[tex]\int {fg'} = fg - \int f'g[/tex]
[tex]\int \ln(u)\ du = \ln(u) * u - \int \frac{1}{u} * u\ du[/tex]
[tex]\int \ln(u)\ du = u\ln(u) - \int \ du[/tex]
So, we have:
[tex]u\ln(u) - \int \ du = u\ln(u) - f(u) + c[/tex]
Integrate du using constant rule
[tex]u\ln(u) - u + c = u\ln(u) - f(u) + c[/tex]
Subtract c from both sides
[tex]u\ln(u) - u = u\ln(u) - f(u)[/tex]
Subtract u ln(u) from both sides
[tex]- u = - f(u)[/tex]
Rewrite as:
[tex]f(u) = u[/tex]
