What is the integral of ((3+ln(x))^3+12)/(4x) ?

The integral of ((3+ln(x))^3+12)/(4x) is 1/4 (((3+ln(x))^4)/4+12ln|x|)+C

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integral of ((3+ln(x))^3+12)/(4x)

\int \frac{( 3 + ln ( x ) ) ^{3} + 12}{4 x} d x

\frac{1}{4} (\frac{( 3 + ln ( x ) ) ^{4}}{4} + 12 ln ∣ x ∣) + C

Solution steps

\int \frac{( 3 + ln ( x ) ) ^{3} + 12}{4 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{( 3 + ln ( x ) ) ^{3} + 12}{x} d x

Expand

\frac{( 3 + ln ( x ) ) ^{3} + 12}{x} : \frac{( 3 + ln ( x ) ) ^{3}}{x} + \frac{12}{x}

= \frac{1}{4} \cdot \int \frac{( 3 + ln ( x ) ) ^{3}}{x} + \frac{12}{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

= \frac{1}{4} (\int \frac{( 3 + ln ( x ) ) ^{3}}{x} d x + \int \frac{12}{x} d x)

\int \frac{( 3 + ln ( x ) ) ^{3}}{x} d x = \frac{( 3 + ln ( x ) ) ^{4}}{4}

\int \frac{12}{x} d x = 12 ln ∣ x ∣

= \frac{1}{4} (\frac{( 3 + ln ( x ) ) ^{4}}{4} + 12 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{4} (\frac{( 3 + ln ( x ) ) ^{4}}{4} + 12 ln ∣ x ∣) + C

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What is the integral of ((3+ln(x))^3+12)/(4x) ?
The integral of ((3+ln(x))^3+12)/(4x) is 1/4 (((3+ln(x))^4)/4+12ln|x|)+C