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

The integral from 1 to 2 of (2x+1)/(x^2+x) is ln(3)

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

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{6}

Compute the boundaries:

ln (6) - ln (2)

= ln (6) - ln (2)

Simplify

= ln (3)

\int_{1}^{3} 4 x^{3} - 8 x + 1 d x

9 t sin (π t)

\frac{1}{( 1 - x ) ^{3}}

\int - 2 x^{2} + 4 d x

\int (- e^{2 - x}) d x

What is the integral from 1 to 2 of (2x+1)/(x^2+x) ?
The integral from 1 to 2 of (2x+1)/(x^2+x) is ln(3)