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

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

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

integral

\frac{u}{( 1 + u ^{2} )}

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

Solution steps

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

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

Add a constant to the solution

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

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

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

tan (2 - 3 x)

e^{x^{2}} x^{2}

3 x^{2} - 9 x + 5

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