What is the integral of 10xsqrt(5x^2-4) ?

The integral of 10xsqrt(5x^2-4) is 2/3 (5x^2-4)^{3/2}+C

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integral of 10xsqrt(5x^2-4)

\int 10 x 5 x^{2} - 4 d x

\frac{2}{3} (5 x^{2} - 4)^{\frac{3}{2}} + C

Solution steps

\int 10 x 5 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 5 x^{2} - 4 d x

Apply u-substitution

: \int \frac{u}{10} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 5 x^{2} - 4

= 10 \cdot \frac{1}{10} \cdot \frac{2}{3} (5 x^{2} - 4)^{\frac{3}{2}}

Simplify

10 \cdot \frac{1}{10} \cdot \frac{2}{3} (5 x^{2} - 4)^{\frac{3}{2}} : \frac{2}{3} (5 x^{2} - 4)^{\frac{3}{2}}

= \frac{2}{3} (5 x^{2} - 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (5 x^{2} - 4)^{\frac{3}{2}} + C

\int_{0}^{x^{3}} sin (t) d t

x \to - 5 + lim (ln (x + 5))

\frac{s + 5}{( s + 3 ) ( s - 1 )}

x \to 9 lim (\frac{- 1}{x - 9})

\int x \cdot 5^{x^{2}} d x

What is the integral of 10xsqrt(5x^2-4) ?
The integral of 10xsqrt(5x^2-4) is 2/3 (5x^2-4)^{3/2}+C