What is the integral of (x^3)/(3x^4-7) ?

The integral of (x^3)/(3x^4-7) is 1/12 ln|3x^4-7|+C

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integral of (x^3)/(3x^4-7)

\int \frac{x ^{3}}{3 x ^{4} - 7} d x

\frac{1}{12} ln 3 x^{4} - 7 + C

Solution steps

\int \frac{x ^{3}}{3 x ^{4} - 7} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= \int \frac{1}{12 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} ln ∣ u ∣

Substitute back

u = 3 x^{4} - 7

= \frac{1}{12} ln 3 x^{4} - 7

Add a constant to the solution

= \frac{1}{12} ln 3 x^{4} - 7 + C

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What is the integral of (x^3)/(3x^4-7) ?
The integral of (x^3)/(3x^4-7) is 1/12 ln|3x^4-7|+C