What is the integral from 0 to 1 of (5x^4)/(1+x^5) ?

The integral from 0 to 1 of (5x^4)/(1+x^5) is ln(2)

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integral from 0 to 1 of (5x^4)/(1+x^5)

\int_{0}^{1} \frac{5 x ^{4}}{1 + x ^{5}} d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{1} \frac{5 x ^{4}}{1 + x ^{5}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{x ^{4}}{1 + x ^{5}} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{5 u} d u

= 5 \cdot \int_{1}^{2} \frac{1}{5 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{5} [ln ∣ u ∣]_{1}^{2}

Simplify

5 \cdot \frac{1}{5} [ln ∣ u ∣]_{1}^{2} : [ln ∣ u ∣]_{1}^{2}

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\int (6 x - 2)^{- 4} d x

\int \frac{2 r}{( 1 - r ) ^{7}} d r

y = x^{2}, y = 2 - x

\int - \frac{2}{1 - 9 x ^{2}} d x

\frac{d y}{d x} = - sin^{2} (y) + 1

What is the integral from 0 to 1 of (5x^4)/(1+x^5) ?
The integral from 0 to 1 of (5x^4)/(1+x^5) is ln(2)