What is the integral of (x^3)/(x^4+7) ?

The integral of (x^3)/(x^4+7) is 1/4 ln|x^4+7|+C

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integral of (x^3)/(x^4+7)

\int \frac{x ^{3}}{x ^{4} + 7} d x

\frac{1}{4} ln x^{4} + 7 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} + 7} 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} + 7

= \frac{1}{4} ln x^{4} + 7

Add a constant to the solution

= \frac{1}{4} ln x^{4} + 7 + C

x \to 1 lim (arctan^{2} (x))

\frac{\partial}{\partial x} ((sin^{2} (x)) + y)

3 t - t^{2}

\frac{d}{d x} (cos (ln (sec (e^{- 2 x}))))

f (x) = (5 x - 5) (x + 3)

What is the integral of (x^3)/(x^4+7) ?
The integral of (x^3)/(x^4+7) is 1/4 ln|x^4+7|+C