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

The integral from 0 to 1 of (x+2)/(x^2+4x+1) is 1/2 ln(6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of (x+2)/(x^2+4x+1)

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

\frac{1}{2} ln (6)

Decimal

0.89587 \dots

Solution steps

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

Complete the square

x^{2} + 4 x + 1 : (x + 2)^{2} - 3

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

Apply u-substitution

= \int_{2}^{3} \frac{u}{u ^{2} - 3} d u

Apply u-substitution

= \int_{1}^{6} \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{6} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} [ln ∣ v ∣]_{1}^{6}

Compute the boundaries:

ln (6)

= \frac{1}{2} ln (6)

What is the integral from 0 to 1 of (x+2)/(x^2+4x+1) ?
The integral from 0 to 1 of (x+2)/(x^2+4x+1) is 1/2 ln(6)