What is the integral from 0 to 1 of (3x^3)/(4x^4+9) ?

The integral from 0 to 1 of (3x^3)/(4x^4+9) is 3/16 (ln(13)-2ln(3))

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integral from 0 to 1 of (3x^3)/(4x^4+9)

\int_{0}^{1} \frac{3 x ^{3}}{4 x ^{4} + 9} d x

\frac{3}{16} (ln (13) - 2 ln (3))

Solution steps

\int_{0}^{1} \frac{3 x ^{3}}{4 x ^{4} + 9} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{x ^{3}}{4 x ^{4} + 9} d x

Apply u-substitution

: \int_{9}^{13} \frac{1}{16 u} d u

= 3 \cdot \int_{9}^{13} \frac{1}{16 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int_{9}^{13} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{16} [ln ∣ u ∣]_{9}^{13}

Simplify

3 \cdot \frac{1}{16} [ln ∣ u ∣]_{9}^{13} : \frac{3}{16} [ln ∣ u ∣]_{9}^{13}

= \frac{3}{16} [ln ∣ u ∣]_{9}^{13}

Compute the boundaries:

ln (13) - 2 ln (3)

= \frac{3}{16} (ln (13) - 2 ln (3))

\frac{d}{d x} (\frac{2}{( 2 - x ^{8} ) ^{\frac{3}{2}}})

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

\frac{1}{4 x}

\int_{4}^{9} \frac{x - x}{x} d x

\int 3 \cdot x + 4 d x

What is the integral from 0 to 1 of (3x^3)/(4x^4+9) ?
The integral from 0 to 1 of (3x^3)/(4x^4+9) is 3/16 (ln(13)-2ln(3))