What is the integral of (9x(6x^2+5)^{1/2})/2 ?

The integral of (9x(6x^2+5)^{1/2})/2 is 1/4 (6x^2+5)^{3/2}+C

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integral of (9x(6x^2+5)^{1/2})/2

\int \frac{9 x ( 6 x ^{2} + 5 ) ^{\frac{1}{2}}}{2} d x

\frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}} + C

Solution steps

\int \frac{9 x ( 6 x ^{2} + 5 ) ^{\frac{1}{2}}}{2} d x

Take the constant out:

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

= \frac{9}{2} \cdot \int x (6 x^{2} + 5)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{( 6 u + 5 ) ^{\frac{1}{2}}}{2} d u

= \frac{9}{2} \cdot \int \frac{( 6 u + 5 ) ^{\frac{1}{2}}}{2} d u

Take the constant out:

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

= \frac{9}{2} \cdot \frac{1}{2} \cdot \int (6 u + 5)^{\frac{1}{2}} d u

Apply u-substitution

: \int \frac{v ^{\frac{1}{2}}}{6} d v

= \frac{9}{2} \cdot \frac{1}{2} \cdot \int \frac{v ^{\frac{1}{2}}}{6} d v

Take the constant out:

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

= \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

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

= \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= \frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{3} (6 x^{2} + 5)^{\frac{3}{2}}

Simplify

\frac{9}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \frac{2}{3} (6 x^{2} + 5)^{\frac{3}{2}} : \frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}}

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

Add a constant to the solution

= \frac{1}{4} (6 x^{2} + 5)^{\frac{3}{2}} + C

\int \frac{1}{19 y} d y

\frac{4}{s ^{2} + 7 s + 12}

x \to 0 lim (e^{x} + 1)

a + b e^{θ}

\frac{d}{d x} ((x^{3} + 2 x^{\frac{1}{2}}) (x^{\frac{1}{3}} + 5))

What is the integral of (9x(6x^2+5)^{1/2})/2 ?
The integral of (9x(6x^2+5)^{1/2})/2 is 1/4 (6x^2+5)^{3/2}+C