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

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

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integral of sqrt(r^2+1)r

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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