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

The integral of xsqrt(4x^2-5) is 1/12 (4x^2-5)^{3/2}+C

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integral of xsqrt(4x^2-5)

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

\frac{1}{12} (4 x^{2} - 5)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

= \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{1}{8} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = 4 x^{2} - 5

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

Simplify

\frac{1}{8} \cdot \frac{2}{3} (4 x^{2} - 5)^{\frac{3}{2}} : \frac{1}{12} (4 x^{2} - 5)^{\frac{3}{2}}

= \frac{1}{12} (4 x^{2} - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (4 x^{2} - 5)^{\frac{3}{2}} + C

f (x) = (3 x^{4} + 8)^{\frac{4}{3}}

\int_{0}^{1} 2 (1 + x)^{4} d x

\frac{d}{d x} (cos (x^{2}) - 2 x^{2} sin (x^{2}))

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

y = 1 - cos^{2} (x)

What is the integral of xsqrt(4x^2-5) ?
The integral of xsqrt(4x^2-5) is 1/12 (4x^2-5)^{3/2}+C