What is the integral of (2x^3)/((x^4+5)) ?

The integral of (2x^3)/((x^4+5)) is 1/2 ln|x^4+5|+C

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integral of (2x^3)/((x^4+5))

\int \frac{2 x ^{3}}{( x ^{4} + 5 )} d x

\frac{1}{2} ln x^{4} + 5 + C

Solution steps

\int \frac{2 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

= 2 \cdot \int \frac{x ^{3}}{x ^{4} + 5} d x

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 5

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

Simplify

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

= \frac{1}{2} ln x^{4} + 5

Add a constant to the solution

= \frac{1}{2} ln x^{4} + 5 + C

1.12 x^{3} - 6.3 x^{2} - 11.2 x + 1.2

x^{2} e^{x}

\frac{d}{d x} (\frac{i + x j + 2 x k}{5 x ^{2} + 1})

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

(x - 5) y^{^{'}} + y = x^{2} - 17

What is the integral of (2x^3)/((x^4+5)) ?
The integral of (2x^3)/((x^4+5)) is 1/2 ln|x^4+5|+C