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

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

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

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

ln ln (3 x^{4} - 4) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{12 u ln ( u )} d u

= 12 \cdot \int \frac{1}{12 u ln ( u )} d u

Take the constant out:

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

= 12 \cdot \frac{1}{12} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

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

= 12 \cdot \frac{1}{12} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 12 \cdot \frac{1}{12} ln ∣ v ∣

Substitute back

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

Simplify

12 \cdot \frac{1}{12} ln ln (3 x^{4} - 4) : ln ln (3 x^{4} - 4)

= ln ln (3 x^{4} - 4)

Add a constant to the solution

= ln ln (3 x^{4} - 4) + C

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