What is the integral of 6xsqrt(3x^2-1) ?

The integral of 6xsqrt(3x^2-1) is 2/3 (3x^2-1)^{3/2}+C

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integral of 6xsqrt(3x^2-1)

\int 6 x 3 x^{2} - 1 d x

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

Solution steps

\int 6 x 3 x^{2} - 1 d x

Take the constant out:

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

= 6 \cdot \int x 3 x^{2} - 1 d x

Apply u-substitution

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

= 6 \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

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

Apply the Power Rule

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

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

Substitute back

u = 3 x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{2}{n \cdot ( n + 1 )}

\int w^{2} d w

g (x) = (4 x^{3} + 6 x^{2})^{5}

\frac{d}{d θ} (2 sin^{2} (θ))

\frac{d}{d x} ((a x - b) (c x^{2}))

What is the integral of 6xsqrt(3x^2-1) ?
The integral of 6xsqrt(3x^2-1) is 2/3 (3x^2-1)^{3/2}+C