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

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

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integral of xsqrt(x^2+9)

\int x 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

\int (e^{x} + 2)^{2} d x

\frac{\partial}{\partial x} (- 5 y^{3} sin (5 x))

y = \frac{2}{x}, y = 8 x, y = \frac{1}{8} x

\int_{2}^{\infty} \frac{1}{ln ( x )} d x

x \to \infty lim (- \frac{1}{2} e^{- x^{2}})

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