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

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

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integral of rsqrt(1+r^2)

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = 1 + r^{2}

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

Simplify

\frac{( 1 + r ^{2} ) ^{3}}{3} : \frac{1}{3} (1 + r^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

3 x^{2} e^{x} + e^{x} (x^{3} + 1)

\int - 4 x d x

\frac{d y}{d x} = \frac{y ( - 2 x + 4 )}{( x ^{2} + 1 ) ( x - 1 ) ^{2}}

3 arccos (\frac{x}{2})

x \to 0 lim (\frac{3 ^{x} - 1 1 ^{x}}{x})

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