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

The integral from 0 to 1 of 1/(x+y) is ln|x+1|-ln|x|

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integral from 0 to 1 of 1/(x+y)

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

ln ∣ x + 1 ∣ - ln ∣ x ∣

Solution steps

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

Apply u-substitution

: \int_{x}^{x + 1} \frac{1}{u} d u

= \int_{x}^{x + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{x}^{x + 1}

Compute the boundaries:

ln ∣ x + 1 ∣ - ln ∣ x ∣

= ln ∣ x + 1 ∣ - ln ∣ x ∣

\int_{\frac{π}{4}}^{\frac{π}{2}} - 2 x csc^{2} (x) d x

\int_{0}^{1} - ln (x) d x

\int_{- 2}^{2} (- 7 x + 2) d x

\int_{0}^{1} x^{\frac{4}{3}} d x

\int_{1}^{2} ln (3) - 2 y d y

What is the integral from 0 to 1 of 1/(x+y) ?
The integral from 0 to 1 of 1/(x+y) is ln|x+1|-ln|x|