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

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

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

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

ln 4 + 5 x^{4} + C

Solution steps

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

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

Simplify

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

= ln 4 + 5 x^{4}

Add a constant to the solution

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

y^{^{''}} + 4 y = t \cdot sin (2 t)

2 + 5 arctan (\frac{x}{2})

\frac{1}{( x + 6 ) ^{2}}

x y^{^{'}} + y = yy

\frac{\partial}{\partial y} (x^{2} - y x)

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