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

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

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

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

x (ln ∣ x + 2 ∣ - ln ∣ x + 1 ∣)

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= x [ln ∣ u ∣]_{x + 1}^{x + 2}

Compute the boundaries:

ln ∣ x + 2 ∣ - ln ∣ x + 1 ∣

= x (ln ∣ x + 2 ∣ - ln ∣ x + 1 ∣)

\int_{0}^{1} (1 + 2 x)^{3} d x

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

\int_{0}^{1} (3 x^{2} + 4 x - \frac{1}{2}) d x

\int_{- 1}^{1} - x^{2} + 9 d x

\int_{0}^{3} \frac{3 x - 2}{x + 1} d x

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