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

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

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integral of x*sqrt(x^2+9)

\int x \cdot x^{2} + 9 d x

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

Solution steps

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

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 = x^{2} + 9

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{( x ^{2} + 1 ) ^{2}}

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