What is the integral of rsqrt(1+4r^2) ?

The integral of rsqrt(1+4r^2) is 1/12 (1+4r^2)^{3/2}+C

integral of rsqrt(1+4r^2)

\int r 1 + 4 r^{2} d r

\frac{1}{12} (1 + 4 r^{2})^{\frac{3}{2}} + C

Solution steps

\int r 1 + 4 r^{2} d r

Apply u-substitution

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 4 r^{2}

= \frac{1}{8} \cdot \frac{2}{3} (1 + 4 r^{2})^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (1 + 4 r^{2})^{\frac{3}{2}} : \frac{1}{12} (1 + 4 r^{2})^{\frac{3}{2}}

= \frac{1}{12} (1 + 4 r^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (1 + 4 r^{2})^{\frac{3}{2}} + C

t an g e n t f (x) = cos (x), at x = \frac{π}{4}

d er i v a t i v e x 5^{x}

t an g e n t \frac{2}{x}, at x = 3

\int \frac{2}{2 x + 1} d x

n = 1 \sum \infty 23 (\frac{2}{5})^{n}

What is the integral of rsqrt(1+4r^2) ?
The integral of rsqrt(1+4r^2) is 1/12 (1+4r^2)^{3/2}+C