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

\int \frac{1}{( x ^{2} - 9 ) ^{2}} d x

\frac{d}{d x} (3 x - 8)

f (x) = (1 + 3 x)^{10}, at x = 0

x \to 2 lim (x^{2} + 4 x)

2 y \frac{d y}{d x} + 2 = y^{2} + 2 x

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