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This question is designed to be answered without a calculator. Use the antiderivative formula shown, where f(u) represents a function. Integral of (l n u) d u = u ln u – f(u) + C Which function represents f(u)?

Sagot :

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]