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

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

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integral of y/(1+x^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 4 + 4 x^{2} - 2 x^{3}, at x = a

t \to 0 lim (sin (6 t))

d y - (2 y - e^{2 x}) d x = 0

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

\frac{d}{d x} (10 x^{3} tan (x))

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