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

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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

f (x) = cos^{4} (sin^{3} (x))

f (x) = x^{4} + 7 x^{2} - x, at x = 1

\frac{( x ^{2} + 1 )}{( x + 1 )}, 1

\frac{\partial}{\partial u} (u - uv)

\frac{2}{x ^{4}}

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