What is the integral of (8x^3)/((x^4+5)) ?

The integral of (8x^3)/((x^4+5)) is 2ln|x^4+5|+C

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integral of (8x^3)/((x^4+5))

\int \frac{8 x ^{3}}{( x ^{4} + 5 )} d x

2 ln x^{4} + 5 + C

Solution steps

\int \frac{8 x ^{3}}{x ^{4} + 5} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{3}}{x ^{4} + 5} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 8 \cdot \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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 5

= 8 \cdot \frac{1}{4} ln x^{4} + 5

Simplify

8 \cdot \frac{1}{4} ln x^{4} + 5 : 2 ln x^{4} + 5

= 2 ln x^{4} + 5

Add a constant to the solution

= 2 ln x^{4} + 5 + C

\int \frac{x ^{3}}{x ^{2} + x - 6} d x

\frac{d}{d x} (\frac{2}{x} - \frac{3}{x ^{2}})

x^{3}, x^{2}

\frac{d}{d x} ((x + cos (x))^{c o t (x)})

y^{^{''}} - 2 y^{^{'}} + y = t e^{t} + 4, y (0) = 1, y^{^{'}} (0) = 1

What is the integral of (8x^3)/((x^4+5)) ?
The integral of (8x^3)/((x^4+5)) is 2ln|x^4+5|+C