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

\frac{d}{d x} (7 x + 9)

n = 0 \sum \infty \frac{1}{n + 6}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x})

\int e^{x} cos (4 x) d x

f (x) = x^{4} - 17 x^{2} + 16, at x = - 2

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