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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Take the constant out:

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

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

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)

y^{^{''}} + 9 y = 0, y (0) = - 6, y^{^{'}} (0) = - 9

\frac{d}{d x} (\frac{- 8 x}{( x ^{2} - 4 ) ^{2}})

\int_{0}^{1} \frac{1}{e ^{x}} d x

e^{x^{2} y}

\frac{\partial}{\partial x} (2 ln (x) + y)

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