What is the integral of r/(sqrt(4+r^2)) ?

The integral of r/(sqrt(4+r^2)) is sqrt(4+r^2)+C

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integral of r/(sqrt(4+r^2))

\int \frac{r}{4 + r ^{2}} d r

4 + r^{2} + C

Solution steps

\int \frac{r}{4 + 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 = 4 + r^{2}

= \frac{1}{2} \cdot 2 (4 + r^{2})^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (4 + r^{2})^{\frac{1}{2}} : 4 + r^{2}

= 4 + r^{2}

Add a constant to the solution

= 4 + r^{2} + C

\int 91 - 6 x - x^{2} d x

y (1 + x^{2}) y^{^{'}} - x (7 + y^{2}) = 0

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

\frac{d}{d x} (\frac{- x ^{2}}{2 x ^{2} - 3})

f^{^{'}} (x) = 7 x^{4} - 9 x^{3} - 4

What is the integral of r/(sqrt(4+r^2)) ?
The integral of r/(sqrt(4+r^2)) is sqrt(4+r^2)+C