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)

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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 Notation

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

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

Apply u-substitution

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

= \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)

\int_{1}^{3} (x^{2} - 4 x + 6) d x

\int_{0}^{10} \frac{t ^{2}}{2} + 4 t + 5 d t

\int_{0}^{3} x x^{2} + 4 d x

\int_{0}^{b} 5 e^{x} d x

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

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)