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

The integral from 0 to 1 of (x^3)/(2(1+x^4)) is 1/8 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 (x^3)/(2(1+x^4))

\int_{0}^{1} \frac{x ^{3}}{2 ( 1 + x ^{4} )} d x

\frac{1}{8} ln (2)

Decimal

0.08664 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3}}{2 ( 1 + x ^{4} )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} \frac{x ^{3}}{1 + x ^{4}} d x

Apply u-substitution

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

= \frac{1}{2} \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

= \frac{1}{2} \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 ∣)

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

Simplify

\frac{1}{2} \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{2} : \frac{1}{8} [ln ∣ u ∣]_{1}^{2}

= \frac{1}{8} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= \frac{1}{8} ln (2)

\int_{- 4}^{4} x^{3} - 16 x d x

\int_{1}^{2} x^{3} - x^{2} d x

\int_{- 2}^{1} x + 2 d x

\int_{1}^{\infty} x \cdot \frac{4}{x ^{5}} d x

\int_{1}^{3} \frac{( x ^{2} + 5 )}{x} d x

What is the integral from 0 to 1 of (x^3)/(2(1+x^4)) ?
The integral from 0 to 1 of (x^3)/(2(1+x^4)) is 1/8 ln(2)