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

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)

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

(0, - 3), (2, 0)

\int_{1}^{3} x 3^{(x^{2})} d x

\frac{d y}{d x} = (\frac{2 y + 5}{4 x + 9})^{2}

\int_{0}^{\infty} \int_{0}^{\infty} e^{- x - y} d x d 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)