What is the integral of 3xsqrt((3x^2+7)) ?

The integral of 3xsqrt((3x^2+7)) is 1/3 (3x^2+7)^{3/2}+C

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integral of 3xsqrt((3x^2+7))

\int 3 x (3 x^{2} + 7) d x

\frac{1}{3} (3 x^{2} + 7)^{\frac{3}{2}} + C

Solution steps

\int 3 x 3 x^{2} + 7 d x

Take the constant out:

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

= 3 \cdot \int x 3 x^{2} + 7 d x

Apply u-substitution

: \int \frac{u}{6} d u

= 3 \cdot \int \frac{u}{6} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x^{2} + 7

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 x^{2} + 7)^{\frac{3}{2}}

Simplify

3 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 x^{2} + 7)^{\frac{3}{2}} : \frac{1}{3} (3 x^{2} + 7)^{\frac{3}{2}}

= \frac{1}{3} (3 x^{2} + 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (3 x^{2} + 7)^{\frac{3}{2}} + C

f (x) = 3 x^{4} + 2 e^{x}, at x = 0

\frac{( s + 3 )}{s ^{2} + s + 3}

\frac{d}{d x} \frac{d y}{d x} - a^{2} y + b = 0

x \to \infty lim (\frac{45}{x})

\frac{d}{d t} (\frac{4}{3} t^{3})

What is the integral of 3xsqrt((3x^2+7)) ?
The integral of 3xsqrt((3x^2+7)) is 1/3 (3x^2+7)^{3/2}+C