What is the integral of (x^3)/(2x^4+1) ?

The integral of (x^3)/(2x^4+1) is 1/8 ln|2x^4+1|+C

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integral of (x^3)/(2x^4+1)

\int \frac{x ^{3}}{2 x ^{4} + 1} d x

\frac{1}{8} ln 2 x^{4} + 1 + C

Solution steps

\int \frac{x ^{3}}{2 x ^{4} + 1} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 2 x^{4} + 1

= \frac{1}{8} ln 2 x^{4} + 1

Add a constant to the solution

= \frac{1}{8} ln 2 x^{4} + 1 + C

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f (x) = 7 x - x^{2}

\frac{d y}{d x}, (e^{x y}) = x + y

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

n = 1 \sum \infty \frac{1}{1 + 2 + n}

What is the integral of (x^3)/(2x^4+1) ?
The integral of (x^3)/(2x^4+1) is 1/8 ln|2x^4+1|+C