What is the integral of sqrt(9+y^2)y ?

The integral of sqrt(9+y^2)y is 1/3 (9+y^2)^{3/2}+C

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integral of sqrt(9+y^2)y

\int 9 + y^{2} y d y

\frac{1}{3} (9 + y^{2})^{\frac{3}{2}} + C

Solution steps

\int 9 + y^{2} y d y

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 = 9 + y^{2}

= \frac{( 9 + y ^{2} ) ^{3}}{3}

Simplify

\frac{( 9 + y ^{2} ) ^{3}}{3} : \frac{1}{3} (9 + y^{2})^{\frac{3}{2}}

= \frac{1}{3} (9 + y^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (9 + y^{2})^{\frac{3}{2}} + C

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What is the integral of sqrt(9+y^2)y ?
The integral of sqrt(9+y^2)y is 1/3 (9+y^2)^{3/2}+C