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

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

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integral of-4/(xln(x^4))

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

- ln (4 ∣ ln (x) ∣) + C

Solution steps

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

Take the constant out:

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

= - 4 \cdot \int \frac{1}{x ln ( x ^{4} )} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - 4 \cdot \frac{1}{4} ln ∣ v ∣

Substitute back

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

Simplify

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

= - ln (4 ∣ ln (x) ∣)

Add a constant to the solution

= - ln (4 ∣ ln (x) ∣) + C

\int \frac{ln ( 2 x )}{x ^{4}} d x

\int \frac{x ^{3} + 1}{x ^{4}} d x

\int (2 x + 1) ln^{2} (x) d x

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

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