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

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

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integral of r/(sqrt(1+r^2))

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

1 + r^{2} + C

Solution steps

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

Apply u-substitution

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

= \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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 + r^{2}

= \frac{1}{2} \cdot 2 (1 + r^{2})^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (1 + r^{2})^{\frac{1}{2}} : 1 + r^{2}

= 1 + r^{2}

Add a constant to the solution

= 1 + r^{2} + C

f (x) = sec (x), at x = \frac{π}{6}

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\frac{\partial}{\partial x} (12 - x^{2} - y^{2})

y^{^{'}} = k y (100 - y)

\frac{d}{d x} (11 sin (x))

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