What is the integral from 0 to 1 of (18x)/(x^2+1) ?

The integral from 0 to 1 of (18x)/(x^2+1) is 9ln(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 (18x)/(x^2+1)

\int_{0}^{1} \frac{18 x}{x ^{2} + 1} d x

9 ln (2)

Decimal Notation

6.23832 \dots

Solution steps

\int_{0}^{1} \frac{18 x}{x ^{2} + 1} d x

Take the constant out:

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

= 18 \cdot \int_{0}^{1} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

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

= 18 \cdot \int_{1}^{2} \frac{1}{2 u} d u

Take the constant out:

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

= 18 \cdot \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 18 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Simplify

18 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : 9 [ln ∣ u ∣]_{1}^{2}

= 9 [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= 9 ln (2)

\int (x^{7} e^{x^{8}}) d x

\int \frac{1}{x ^{2} ( x + 1 )} d x

\int \frac{a + x}{a - x} d x

f (x) = \frac{1}{3 - x}

n = 1 \sum \infty \frac{n}{( n ^{2} + 1 )}

What is the integral from 0 to 1 of (18x)/(x^2+1) ?
The integral from 0 to 1 of (18x)/(x^2+1) is 9ln(2)