What is the integral of 10xsqrt(3x^2-4) ?

The integral of 10xsqrt(3x^2-4) is 10/9 (3x^2-4)^{3/2}+C

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integral of 10xsqrt(3x^2-4)

\int 10 x 3 x^{2} - 4 d x

\frac{10}{9} (3 x^{2} - 4)^{\frac{3}{2}} + C

Solution steps

\int 10 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

= 10 \cdot \int x 3 x^{2} - 4 d x

Apply u-substitution

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

= 10 \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

= 10 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

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

= 10 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x^{2} - 4

= 10 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 x^{2} - 4)^{\frac{3}{2}}

Simplify

10 \cdot \frac{1}{6} \cdot \frac{2}{3} (3 x^{2} - 4)^{\frac{3}{2}} : \frac{10}{9} (3 x^{2} - 4)^{\frac{3}{2}}

= \frac{10}{9} (3 x^{2} - 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{10}{9} (3 x^{2} - 4)^{\frac{3}{2}} + C

\frac{\partial}{\partial y} (x^{2} + 2 y^{2} - x^{2} y)

x \to \frac{7 π}{2} + lim (sec (x))

\frac{d}{d x} ((2 x - 9)^{4} (x^{2} + x + 1)^{5})

\frac{d}{d x} (3 ln (15 x))

x \to 3 lim (cos (\frac{π}{4} x))

What is the integral of 10xsqrt(3x^2-4) ?
The integral of 10xsqrt(3x^2-4) is 10/9 (3x^2-4)^{3/2}+C