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

The integral from 1 to 2 of 1/(x+y) is ln|y+2|-ln|y+1|

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

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

ln ∣ y + 2 ∣ - ln ∣ y + 1 ∣

Solution steps

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

Apply u-substitution

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

= \int_{y + 1}^{y + 2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{y + 1}^{y + 2}

Compute the boundaries:

ln ∣ y + 2 ∣ - ln ∣ y + 1 ∣

= ln ∣ y + 2 ∣ - ln ∣ y + 1 ∣

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

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

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

\int_{2}^{3} (3 t + 2) d t

What is the integral from 1 to 2 of 1/(x+y) ?
The integral from 1 to 2 of 1/(x+y) is ln|y+2|-ln|y+1|