What is the integral of 1/(5+2x^4)(8x^3) ?

The integral of 1/(5+2x^4)(8x^3) is ln|5+2x^4|+C

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integral of 1/(5+2x^4)(8x^3)

\int \frac{1}{5 + 2 x ^{4}} (8 x^{3}) d x

ln 5 + 2 x^{4} + C

Solution steps

\int \frac{1}{5 + 2 x ^{4}} \cdot 8 x^{3} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{5 + 2 x ^{4}} x^{3} d x

Simplify

\frac{1}{5 + 2 x ^{4}} x^{3} : \frac{x ^{3}}{5 + 2 x ^{4}}

= 8 \cdot \int \frac{x ^{3}}{5 + 2 x ^{4}} d x

Apply u-substitution

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

= 8 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 5 + 2 x^{4}

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

Simplify

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

= ln 5 + 2 x^{4}

Add a constant to the solution

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

\frac{d}{d x} (8 sin (x) - 4 2 x)

{cos (t)}

f (t) = cos (8 t)

f (x) = x^{3} - 3 x + 4

What is the integral of 1/(5+2x^4)(8x^3) ?
The integral of 1/(5+2x^4)(8x^3) is ln|5+2x^4|+C