What is the integral of 3xsqrt(3x^2+4) ?

The integral of 3xsqrt(3x^2+4) is 1/3 (3x^2+4)^{3/2}+C

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integral of 3xsqrt(3x^2+4)

\int 3 x 3 x^{2} + 4 d x

\frac{1}{3} (3 x^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int 3 x 3 x^{2} + 4 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} + 4 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} + 4

= 3 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 x^{2} + 4)^{\frac{3}{2}}

Simplify

3 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 x^{2} + 4)^{\frac{3}{2}} : \frac{1}{3} (3 x^{2} + 4)^{\frac{3}{2}}

= \frac{1}{3} (3 x^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (3 x^{2} + 4)^{\frac{3}{2}} + C

\int \frac{( 2 x )}{( x - 2 ) ( x + 1 )} d x

\frac{\partial}{\partial x} (x \cdot e^{x^{2} - 4})

\frac{d}{d x} ((ln (6 x))^{2})

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

f (t) = \frac{2 t}{2 + t}

What is the integral of 3xsqrt(3x^2+4) ?
The integral of 3xsqrt(3x^2+4) is 1/3 (3x^2+4)^{3/2}+C