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

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

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

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

4 x + C

Solution steps

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

Apply u-substitution

: \int 4 e^{u} d u

= \int 4 e^{u} d u

Take the constant out:

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

= 4 \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 4 e^{u}

Substitute back

u = ln (x)

= 4 e^{l n (x)}

Apply log rule:

a^{l o g_{a} (b)} = b

= 4 x

Add a constant to the solution

= 4 x + C

\int e^{- \frac{1}{3} x} sin (2 x) d x

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\int \frac{5 x}{( 3 - x ^{2} ) ^{3}} d x

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