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

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

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integral of u/(u^2+1)

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

\frac{1}{2} ln u^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

v = u^{2} + 1

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

Add a constant to the solution

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

\int \frac{ln ( x ) + 1}{x} d x

\int ln (x^{2}) d x

\int 2 x (x^{2} + 2)^{- 3} d x

i = 1 \sum \infty 8 (\frac{5}{6})^{i - 1}

(cos (x))^{2} π

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