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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d}{d x} (⌊ \frac{3}{2} ⌋)

\frac{d}{d x} (arccos (\frac{7}{x}))

y = x 4^{x + 9}

x \to \infty lim (- 2 e^{- x})

\int \frac{4}{3 x ^{2} 5 - 7 x ^{2}} d x

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