What is the integral of (x^3)/(x^4-6) ?

The integral of (x^3)/(x^4-6) is 1/4 ln|x^4-6|+C

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integral of (x^3)/(x^4-6)

\int \frac{x ^{3}}{x ^{4} - 6} d x

\frac{1}{4} ln x^{4} - 6 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} - 6} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 6

= \frac{1}{4} ln x^{4} - 6

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 6 + C

\frac{( 2 s + 3 )}{s ^{2} - 7}

\int sec (2 θ) tan (2 θ) d θ

x \to 5 - lim (\frac{- 1}{x ^{2} - 25})

q \to + (- 11) + lim (ln (q + 11))

\int \frac{cos ( 2 θ )}{2} d θ

What is the integral of (x^3)/(x^4-6) ?
The integral of (x^3)/(x^4-6) is 1/4 ln|x^4-6|+C