What is the integral of (20x^4)/(4x^5+3) ?

The integral of (20x^4)/(4x^5+3) is ln|4x^5+3|+C

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integral of (20x^4)/(4x^5+3)

\int \frac{20 x ^{4}}{4 x ^{5} + 3} d x

ln 4 x^{5} + 3 + C

Solution steps

\int \frac{20 x ^{4}}{4 x ^{5} + 3} d x

Take the constant out:

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

= 20 \cdot \int \frac{x ^{4}}{4 x ^{5} + 3} d x

Apply u-substitution

: \int \frac{1}{20 u} d u

= 20 \cdot \int \frac{1}{20 u} d u

Take the constant out:

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

= 20 \cdot \frac{1}{20} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 20 \cdot \frac{1}{20} ln ∣ u ∣

Substitute back

u = 4 x^{5} + 3

= 20 \cdot \frac{1}{20} ln 4 x^{5} + 3

Simplify

20 \cdot \frac{1}{20} ln 4 x^{5} + 3 : ln 4 x^{5} + 3

= ln 4 x^{5} + 3

Add a constant to the solution

= ln 4 x^{5} + 3 + C

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

\frac{d}{d x} (\frac{x ^{2} - 3 x}{x + 1})

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

t \frac{d y}{d t} = t^{3} + 15 t^{3} y

What is the integral of (20x^4)/(4x^5+3) ?
The integral of (20x^4)/(4x^5+3) is ln|4x^5+3|+C