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

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

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

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d y = y

\int 2 y d y = y^{2}

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

Add a constant to the solution

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

\int \frac{1}{3 x} d x

\frac{d}{d x} (cos (x^{s i n (x)}))

\frac{d}{d x} (x^{4} - 50 x^{2})

\frac{\partial}{\partial x} (2 x^{2} + 2 y - 8)

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