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

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

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integral of r/(1+r^2)

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

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

Solution steps

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

Apply u-substitution

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + r^{2}

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

Add a constant to the solution

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

\int \frac{e ^{x}}{1 + 49 e ^{2 x}} d x

\frac{\partial}{\partial y} (x sin (z))

\frac{\partial}{\partial y} (x^{2} - 2 y^{2})

\frac{d}{d x} (\frac{x}{a})

t an g e n t f (x) = (1 + 4 x) (5 - 2 x), at x = 2

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