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

The integral of (2y)/(x^2+1) is (y^2)/(1+x^2)+C

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integral of (2y)/(x^2+1)

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

\frac{y ^{2}}{1 + x ^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{y ^{2}}{2}

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

Simplify

\frac{2}{1 + x ^{2}} \cdot \frac{y ^{2}}{2} : \frac{y ^{2}}{1 + x ^{2}}

= \frac{y ^{2}}{1 + x ^{2}}

Add a constant to the solution

= \frac{y ^{2}}{1 + x ^{2}} + C

\int x + y + z d x

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

\frac{d y}{d x} = \frac{3 y}{x}, x > 0

\int \frac{3}{x ln ( x )} d x

\frac{d}{d x} (\frac{x ^{3}}{39} + \frac{13}{4 x})

What is the integral of (2y)/(x^2+1) ?
The integral of (2y)/(x^2+1) is (y^2)/(1+x^2)+C