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

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

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integral of sqrt(x)(2+xsqrt(x))^5

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

\frac{1}{9} (2 + x^{\frac{3}{2}})^{6} + C

Solution steps

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

Apply u-substitution

: \int 2 u^{2} (2 + u^{3})^{5} d u

= \int 2 u^{2} (2 + u^{3})^{5} d u

Take the constant out:

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

= 2 \cdot \int u^{2} (2 + u^{3})^{5} d u

Apply u-substitution

: \int \frac{v ^{5}}{3} d v

= 2 \cdot \int \frac{v ^{5}}{3} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (- e^{- x} x^{2} + e^{- x} x)

e^{- x}, 5

\frac{d y}{d x} = x y

h (x) = \frac{1}{( - x ^{2} + x - 2 ) ^{5}}

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

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