What is the integral of ((sqrt(x)+2)^4)/(2sqrt(x)) ?

The integral of ((sqrt(x)+2)^4)/(2sqrt(x)) is 1/5 (x^{5/2}+10x^2+40x^{3/2}+80x+80sqrt(x))+C

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integral of ((sqrt(x)+2)^4)/(2sqrt(x))

\int \frac{( x + 2 ) ^{4}}{2 x} d x

\frac{1}{5} (x^{\frac{5}{2}} + 10 x^{2} + 40 x^{\frac{3}{2}} + 80 x + 80 x) + C

Solution steps

\int \frac{( x + 2 ) ^{4}}{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 + 2 ) ^{4}}{x} d x

Long division

\frac{( x + 2 ) ^{4}}{x} : (x)^{3} + 8 x + 24 x + 32 + \frac{16}{x}

= \frac{1}{2} \cdot \int (x)^{3} + 8 x + 24 x + 32 + \frac{16}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int (x)^{3} d x + \int 8 x d x + \int 24 x d x + \int 32 d x + \int \frac{16}{x} d x)

\int (x)^{3} d x = \frac{2}{5} x^{\frac{5}{2}}

\int 8 x d x = 4 x^{2}

\int 24 x d x = 16 x^{\frac{3}{2}}

\int 32 d x = 32 x

\int \frac{16}{x} d x = 32 x

= \frac{1}{2} (\frac{2}{5} x^{\frac{5}{2}} + 4 x^{2} + 16 x^{\frac{3}{2}} + 32 x + 32 x)

Simplify

\frac{1}{2} (\frac{2}{5} x^{\frac{5}{2}} + 4 x^{2} + 16 x^{\frac{3}{2}} + 32 x + 32 x) : \frac{1}{5} (x^{\frac{5}{2}} + 10 x^{2} + 40 x^{\frac{3}{2}} + 80 x + 80 x)

= \frac{1}{5} (x^{\frac{5}{2}} + 10 x^{2} + 40 x^{\frac{3}{2}} + 80 x + 80 x)

Add a constant to the solution

= \frac{1}{5} (x^{\frac{5}{2}} + 10 x^{2} + 40 x^{\frac{3}{2}} + 80 x + 80 x) + C

What is the integral of ((sqrt(x)+2)^4)/(2sqrt(x)) ?
The integral of ((sqrt(x)+2)^4)/(2sqrt(x)) is 1/5 (x^{5/2}+10x^2+40x^{3/2}+80x+80sqrt(x))+C