What is the integral of (x^3)/(x^4+1) ?

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

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integral of (x^3)/(x^4+1)

\int \frac{x ^{3}}{x ^{4} + 1} d x

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

Solution steps

\int \frac{x ^{3}}{x ^{4} + 1} 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} + 1

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (e^{4 x y})

n = 0 \sum \infty (\frac{e}{10})^{n}

y = 5 (x^{3} - x), y = 0

x^{2} (x + 1)^{3}

\frac{d x}{d t} = 6 - \frac{3 x}{500}, x (0) = 0

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