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

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

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

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

ln (2)

Decimal

0.69314 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{2} \frac{1}{4 u} d u

= 4 \cdot \int_{1}^{2} \frac{1}{4 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{2}

Simplify

4 \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{2} : [ln ∣ u ∣]_{1}^{2}

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

8 y^{^{''}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = 7

\int - 4 y d y

f^{^{'}} (x) = sin (x)

\int_{2}^{\infty} x e^{- 3 x} d x

\frac{d}{d t} (5 t^{2} x)

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