What is the integral of 24sqrt(3x+5) ?

The integral of 24sqrt(3x+5) is 16/3 (3x+5)^{3/2}+C

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integral of 24sqrt(3x+5)

\int 24 3 x + 5 d x

\frac{16}{3} (3 x + 5)^{\frac{3}{2}} + C

Solution steps

\int 24 3 x + 5 d x

Take the constant out:

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

= 24 \cdot \int 3 x + 5 d x

Apply u-substitution

: \int \frac{u}{3} d u

= 24 \cdot \int \frac{u}{3} d u

Take the constant out:

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

= 24 \cdot \frac{1}{3} \cdot \int u d u

Apply the Power Rule

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

= 24 \cdot \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x + 5

= 24 \cdot \frac{1}{3} \cdot \frac{2}{3} (3 x + 5)^{\frac{3}{2}}

Simplify

24 \cdot \frac{1}{3} \cdot \frac{2}{3} (3 x + 5)^{\frac{3}{2}} : \frac{16}{3} (3 x + 5)^{\frac{3}{2}}

= \frac{16}{3} (3 x + 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{16}{3} (3 x + 5)^{\frac{3}{2}} + C

\int cos^{2} (X) d X

x \to \infty lim (- sin (x))

x y^{^{'}} + (1 + x) y = e^{- x} sin (2 x)

\frac{d}{d x} (x e^{- 7 x})

f (x) = [(4 x^{2} + 3 x) cos (x)]

What is the integral of 24sqrt(3x+5) ?
The integral of 24sqrt(3x+5) is 16/3 (3x+5)^{3/2}+C