What is the integral of (sqrt(x)+4x-5) ?

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

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integral of (sqrt(x)+4x-5)

\int (x + 4 x - 5) d x

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

Solution steps

\int x + 4 x - 5 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 x d x + \int 4 x d x - \int 5 d x

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

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