What is the integral of 6sqrt(4x-5) ?

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

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integral of 6sqrt(4x-5)

\int 6 4 x - 5 d x

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

Solution steps

\int 6 4 x - 5 d x

Take the constant out:

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

= 6 \cdot \int 4 x - 5 d x

Apply u-substitution

: \int \frac{u}{4} d u

= 6 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x - 5

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

Simplify

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

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

Add a constant to the solution

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

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