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

The integral of (y+1)/(y^2+1) is 1/2 ln|y^2+1|+arctan(y)+C

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

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

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

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

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

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

Add a constant to the solution

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

t an g e n t f (x) = 5 x^{3} - 5 x, (- 1, 0)

\int x (4 x + 5)^{3} d x

im pl i c i t \frac{d y}{d x}, x y = 28

s l o p e (- 1.2) (- 2.3)

\int sin^{4} (y) cos (y) d y

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