What is the integral of ((sqrt(x)+6)^5)/(2sqrt(x)) ?

The integral of ((sqrt(x)+6)^5)/(2sqrt(x)) is 1/6 (sqrt(x)+6)^6+C

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integral of ((sqrt(x)+6)^5)/(2sqrt(x))

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

\frac{1}{6} (x + 6)^{6} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int 2 u^{5} d u

= \frac{1}{2} \cdot \int 2 u^{5} d u

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

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

Substitute back

u = x + 6

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

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{( x + 6 ) ^{6}}{6} : \frac{1}{6} (x + 6)^{6}

= \frac{1}{6} (x + 6)^{6}

Add a constant to the solution

= \frac{1}{6} (x + 6)^{6} + C

\int \frac{1}{x ( ln ( x ) ) ^{9}} d x

\frac{\partial}{\partial x} (x^{2} - 2 x y + 3 y^{2} - 8 y)

\frac{d}{d x} ((f (x))^{3})

\frac{\partial}{\partial x} (ln (9 x^{2} + y^{2} + 9))

x \to 4 lim (x^{2} - 6 x + 8)

What is the integral of ((sqrt(x)+6)^5)/(2sqrt(x)) ?
The integral of ((sqrt(x)+6)^5)/(2sqrt(x)) is 1/6 (sqrt(x)+6)^6+C