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

The integral from 0 to 1 of (x+1)/(x^2+2x+2) is 1/2 (ln(5)-ln(2))

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

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

\frac{1}{2} (ln (5) - ln (2))

Decimal Notation

0.45814 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{5} \frac{1}{2 u} d u

= \int_{2}^{5} \frac{1}{2 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 \int_{2}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

= \frac{1}{2} (ln (5) - ln (2))

\int_{0}^{1} 2 e^{- t} d t

\int_{- 1}^{7} \frac{x ^{2}}{x ^{3} + 1} d x

\int_{- 2}^{0} 4 - 2^{- x} d x

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

\int_{0}^{1} 81 x^{2} + 729 d x

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