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

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

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

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

\frac{ln ( 5 ) x}{2}

Solution steps

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

Take the constant out:

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

= x \cdot \int_{0}^{2 x} \frac{y}{x ^{2} + y ^{2}} d y

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= x \frac{1}{2} [ln ∣ u ∣]_{x^{2}}^{5 x^{2}}

Compute the boundaries:

ln (5)

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

Simplify

= \frac{ln ( 5 ) x}{2}

\int_{- 3}^{0} x^{2} d x

x \to 3 lim (+ (\frac{- 3}{x - 3}))

\frac{d ^{2}}{d x ^{2}} (- 9.8 x^{2} + 2 x + 1)

e^{t} (7 y - 5 t) d t + (6 + 7 e^{t}) d y = 0

\frac{d}{d x} (ln (x^{n}))

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