What is the integral of (3xsqrt(4x^2-6))/5 ?

The integral of (3xsqrt(4x^2-6))/5 is 1/20 (4x^2-6)^{3/2}+C

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integral of (3xsqrt(4x^2-6))/5

\int \frac{3 x 4 x ^{2} - 6}{5} d x

\frac{1}{20} (4 x^{2} - 6)^{\frac{3}{2}} + C

Solution steps

\int \frac{3 x 4 x ^{2} - 6}{5} d x

Take the constant out:

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

= \frac{3}{5} \cdot \int x 4 x^{2} - 6 d x

Apply u-substitution

: \int \frac{u}{8} d u

= \frac{3}{5} \cdot \int \frac{u}{8} d u

Take the constant out:

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

= \frac{3}{5} \cdot \frac{1}{8} \cdot \int u d u

Apply the Power Rule

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

= \frac{3}{5} \cdot \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x^{2} - 6

= \frac{3}{5} \cdot \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} - 6)^{\frac{3}{2}}

Simplify

\frac{3}{5} \cdot \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} - 6)^{\frac{3}{2}} : \frac{1}{20} (4 x^{2} - 6)^{\frac{3}{2}}

= \frac{1}{20} (4 x^{2} - 6)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{20} (4 x^{2} - 6)^{\frac{3}{2}} + C

\int_{1}^{\infty} \frac{y}{y ^{4} + 1} d y

e^{9 x}

f (t) = sin^{2} (t)

\frac{x ^{3}}{2}

\int 8 d x

What is the integral of (3xsqrt(4x^2-6))/5 ?
The integral of (3xsqrt(4x^2-6))/5 is 1/20 (4x^2-6)^{3/2}+C