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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{a} d a

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

f (t) = t^{2} + 5

f (x) = \frac{- 3 x ^{3} - 1}{2 x ^{2} + 1}

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