What is the integral of (3x^2+4x+5)/(sqrt(x)) ?

The integral of (3x^2+4x+5)/(sqrt(x)) is 6/5 x^{5/2}+8/3 x^{3/2}+10sqrt(x)+C

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integral of (3x^2+4x+5)/(sqrt(x))

\int \frac{3 x ^{2} + 4 x + 5}{x} d x

\frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + 10 x + C

Solution steps

\int \frac{3 x ^{2} + 4 x + 5}{x} d x

Expand

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

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

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

\int 3 x^{\frac{3}{2}} d x = \frac{6}{5} x^{\frac{5}{2}}

\int 4 x d x = \frac{8}{3} x^{\frac{3}{2}}

\int \frac{5}{x} d x = 10 x

= \frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + 10 x

Add a constant to the solution

= \frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + 10 x + C

\int 40 arctan (x) d x

\frac{\partial}{\partial y} (16 - 4 x^{2} - y^{2})

\int sec^{2} (5 t) d t

g (x) = x^{2} (1 - 5 x)

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

What is the integral of (3x^2+4x+5)/(sqrt(x)) ?
The integral of (3x^2+4x+5)/(sqrt(x)) is 6/5 x^{5/2}+8/3 x^{3/2}+10sqrt(x)+C