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

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

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

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

\frac{4}{3} ln x^{3} + 4 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= 4 \cdot \int \frac{1}{3 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 4

= 4 \cdot \frac{1}{3} ln x^{3} + 4

Simplify

4 \cdot \frac{1}{3} ln x^{3} + 4 : \frac{4}{3} ln x^{3} + 4

= \frac{4}{3} ln x^{3} + 4

Add a constant to the solution

= \frac{4}{3} ln x^{3} + 4 + C

\int t^{2} e^{- t} d t

x \to 2 lim (\frac{1}{x ^{2} - 1})

y = \frac{1}{8 x ^{2}}, y = x, y = \frac{x}{8}

\frac{d}{d x} (x sec (y))

\frac{1}{x - 4}

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