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

The integral of 4x+3/(sqrt(x)) is 2x^2+6sqrt(x)+C

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integral of 4x+3/(sqrt(x))

\int 4 x + \frac{3}{x} d x

2 x^{2} + 6 x + C

Solution steps

\int 4 x + \frac{3}{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 4 x d x + \int \frac{3}{x} d x

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

\int \frac{3}{x} d x = 6 x

= 2 x^{2} + 6 x

Add a constant to the solution

= 2 x^{2} + 6 x + C

cos (2 t)

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

x^{2} y^{^{'}} - x = 1 + y + x y

\int_{- 1}^{3} (5 - \frac{1}{4} x^{3}) d x

\int \frac{7 x ^{2}}{1 + x ^{6}} d x

What is the integral of 4x+3/(sqrt(x)) ?
The integral of 4x+3/(sqrt(x)) is 2x^2+6sqrt(x)+C