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

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

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integral of 2xsqrt(x^2+9)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int u^{2} d u

= 2 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 2 \cdot \frac{u ^{3}}{3}

Substitute back

u = x^{2} + 9

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((2 x^{2} - 4 x)^{2})

\frac{3}{s s ^{2} + 8 ss + 18 s}

\frac{d}{d x} (4 x - 3 x^{2} - y^{2} - (z - 1)^{2})

\frac{3}{x} - e^{4}

\frac{d}{d x} (- 5 x (2 x - 3)^{4})

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