What is the integral of 1/(2+4x^4)(16x^3) ?

The integral of 1/(2+4x^4)(16x^3) is ln|2+4x^4|+C

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integral of 1/(2+4x^4)(16x^3)

\int \frac{1}{2 + 4 x ^{4}} (16 x^{3}) d x

ln 2 + 4 x^{4} + C

Solution steps

\int \frac{1}{2 + 4 x ^{4}} \cdot 16 x^{3} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{2 + 4 x ^{4}} x^{3} d x

Simplify

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

= 16 \cdot \int \frac{x ^{3}}{2 + 4 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{16 u} d u

= 16 \cdot \int \frac{1}{16 u} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 \cdot \frac{1}{16} ln ∣ u ∣

Substitute back

u = 2 + 4 x^{4}

= 16 \cdot \frac{1}{16} ln 2 + 4 x^{4}

Simplify

16 \cdot \frac{1}{16} ln 2 + 4 x^{4} : ln 2 + 4 x^{4}

= ln 2 + 4 x^{4}

Add a constant to the solution

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

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

\int \frac{x ^{2}}{( x ^{2} - 16 ) ^{\frac{5}{2}}} d x

\int (1 + cos (4 x))^{2} d x

\int v^{3} (a^{2} - v^{2})^{\frac{1}{2}} d v

\int \frac{2 + x ^{2}}{x} d x

What is the integral of 1/(2+4x^4)(16x^3) ?
The integral of 1/(2+4x^4)(16x^3) is ln|2+4x^4|+C