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

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

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integral of sqrt(1+4r^2)r

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

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

Solution steps

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

Apply u-substitution

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

= \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{2}{3} u^{\frac{3}{2}}

= \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

\int \frac{3}{e ^{x}} d x

f (x) = 2^{x^{2}}

\int_{0}^{4} π (\frac{y ^{2}}{2})^{2} d y

\int_{3}^{5} \frac{1}{4 + ( x - 3 ) ^{2}} d x

y^{^{'}} + 3 y = e^{x} y^{2}

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