What is the integral of (ln(12x)-ln(4x)) ?

The integral of (ln(12x)-ln(4x)) is xln(12x)-xln(4x)+C

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integral of (ln(12x)-ln(4x))

\int (ln (12 x) - ln (4 x)) d x

x ln (12 x) - x ln (4 x) + C

Solution steps

\int ln (12 x) - ln (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int ln (12 x) d x - \int ln (4 x) d x

\int ln (12 x) d x = x ln (12 x) - x

\int ln (4 x) d x = x ln (4 x) - x

= x ln (12 x) - x - (x ln (4 x) - x)

Simplify

= x ln (12 x) - x ln (4 x)

Add a constant to the solution

= x ln (12 x) - x ln (4 x) + C

f (x) = \frac{- 8}{x - 1}

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

y^{^{''}} + 5 y^{^{'}} = 0

- \frac{1}{( 1 + x ) ^{2}}

f (x) = x^{5}, at x = 1

What is the integral of (ln(12x)-ln(4x)) ?
The integral of (ln(12x)-ln(4x)) is xln(12x)-xln(4x)+C