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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{sec ^{2} ( x )}{3 + tan ^{2} ( x )} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = t, y (0) = 1, y^{^{'}} (0) = 0

\int_{0}^{1} \frac{x - 8}{x ^{2} - 5 x + 6} d x

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

\frac{d}{d x} (x^{\frac{3}{5}} x^{3} + 2 x - 6)

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