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

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

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

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

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

Solution steps

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

Expand

\frac{u + 1}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{1}{u ^{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{u}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u

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

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

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

Add a constant to the solution

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

d er i v a t i v e f (x) = \frac{e ^{x}}{1 + e ^{x}}

\frac{d}{d x} (tan (x) - 0.5 x)

\frac{d y}{d x} = x y - 3 x - 2 y + 6

\int 5 x ln (x) d x

t an g e n t f (x) = - \frac{1}{x}, at x = 3

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