What is the integral of rsqrt(16+r^2) ?

The integral of rsqrt(16+r^2) is 1/3 (16+r^2)^{3/2}+C

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integral of rsqrt(16+r^2)

\int r 16 + r^{2} d r

\frac{1}{3} (16 + r^{2})^{\frac{3}{2}} + C

Solution steps

\int r 16 + 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 = 16 + r^{2}

= \frac{( 16 + r ^{2} ) ^{3}}{3}

Simplify

\frac{( 16 + r ^{2} ) ^{3}}{3} : \frac{1}{3} (16 + r^{2})^{\frac{3}{2}}

= \frac{1}{3} (16 + r^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (16 + r^{2})^{\frac{3}{2}} + C

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What is the integral of rsqrt(16+r^2) ?
The integral of rsqrt(16+r^2) is 1/3 (16+r^2)^{3/2}+C