What is the integral from 0 to 1 of (2x^4)/(3x^5+20) ?

The integral from 0 to 1 of (2x^4)/(3x^5+20) is 2/15 (ln(23)-ln(20))

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 (2x^4)/(3x^5+20)

\int_{0}^{1} \frac{2 x ^{4}}{3 x ^{5} + 20} d x

\frac{2}{15} (ln (23) - ln (20))

Decimal

0.01863 \dots

Solution steps

\int_{0}^{1} \frac{2 x ^{4}}{3 x ^{5} + 20} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{x ^{4}}{3 x ^{5} + 20} d x

Apply u-substitution

: \int_{20}^{23} \frac{1}{15 u} d u

= 2 \cdot \int_{20}^{23} \frac{1}{15 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{15} \cdot \int_{20}^{23} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{15} [ln ∣ u ∣]_{20}^{23}

Simplify

2 \cdot \frac{1}{15} [ln ∣ u ∣]_{20}^{23} : \frac{2}{15} [ln ∣ u ∣]_{20}^{23}

= \frac{2}{15} [ln ∣ u ∣]_{20}^{23}

Compute the boundaries:

ln (23) - ln (20)

= \frac{2}{15} (ln (23) - ln (20))

\int_{0}^{π} v sin (v) d v

\int_{1}^{2} 6 d x

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

\int_{0}^{1} (2 x - \frac{x}{2}) d x

\int_{- 8}^{8} 1 d x

What is the integral from 0 to 1 of (2x^4)/(3x^5+20) ?
The integral from 0 to 1 of (2x^4)/(3x^5+20) is 2/15 (ln(23)-ln(20))