What is the integral of ((1+x)^2)/(sqrt(x)) ?

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

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integral of ((1+x)^2)/(sqrt(x))

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

2 x + \frac{4}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

Solution steps

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

Expand

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

= \int \frac{1}{x} + 2 x + x^{\frac{3}{2}} 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

= \int \frac{1}{x} d x + \int 2 x d x + \int x^{\frac{3}{2}} d x

\int \frac{1}{x} d x = 2 x

\int 2 x d x = \frac{4}{3} x^{\frac{3}{2}}

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

= 2 x + \frac{4}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}}

Add a constant to the solution

= 2 x + \frac{4}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} + C

\int + \frac{1}{x ^{2} + a ^{2}} d x

\frac{d y}{d x} = 4 y e^{x + 4}

(x + y) d y = (x - y) d x

f (x) = x^{4} - 13 x^{2} + 36, at x = - 2

\frac{d}{d x} (5 x^{3} - 5 x)

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