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

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

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integral of 1+sqrt(x)

\int 1 + x d x

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

Solution steps

\int 1 + 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 1 d x + \int x d x

\int 1 d x = x

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

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

Add a constant to the solution

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

f (x) = 1 - x^{2}

\int y (1 + 3 + y^{2}) d y

n = 1 \sum \infty 14

8 - x

\frac{d}{d v} (u cos (v))

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