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

The integral of (ln(x^4))/x is 2ln^2(x)+C

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integral of (ln(x^4))/x

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

2 ln^{2} (x) + C

Solution steps

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

Apply u-substitution

= \int \frac{ln ( u )}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

= \frac{1}{4} \cdot \int v d v

Apply the Power Rule

= \frac{1}{4} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{4} \cdot \frac{ln ^{2} ( x ^{4} )}{2}

Simplify

\frac{1}{4} \cdot \frac{ln ^{2} ( x ^{4} )}{2} : 2 ln^{2} (x)

= 2 ln^{2} (x)

Add a constant to the solution

= 2 ln^{2} (x) + C

\frac{d}{d x} (\frac{1}{x} e^{x})

\frac{d y}{d x} = a \cdot y

(e^{f (x)})^{'}

\frac{d}{d x} ((\frac{x ^{2} + 2 x}{x + 1})^{5})

2 y^{^{'}} = 3 x y + 1