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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (3t^2)/(1+t^3)

\int_{0}^{1} \frac{3 t ^{2}}{1 + t ^{3}} d t

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{1} \frac{3 t ^{2}}{1 + t ^{3}} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{t ^{2}}{1 + t ^{3}} d t

Apply u-substitution

: \int_{1}^{2} \frac{1}{3 u} d u

= 3 \cdot \int_{1}^{2} \frac{1}{3 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}{3} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{2}

Simplify

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{2} : [ln ∣ u ∣]_{1}^{2}

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

x \to 0 lim (e^{- 2 x})

cos (\frac{x}{3})

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

n = 4 \sum \infty \frac{1}{( n - 4 ) !}

\frac{d}{d t} (t^{2} + \frac{2 t}{t} - \frac{1}{t ^{2}})

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