What is the integral of 1/4 (2x+1) ?

The integral of 1/4 (2x+1) is 1/4 (x^2+x)+C

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integral of 1/4 (2x+1)

\int \frac{1}{4} (2 x + 1) d x

\frac{1}{4} (x^{2} + x) + C

Solution steps

\int \frac{1}{4} (2 x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int 2 x + 1 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

= \frac{1}{4} (\int 2 x d x + \int 1 d x)

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

\int 1 d x = x

= \frac{1}{4} (x^{2} + x)

Add a constant to the solution

= \frac{1}{4} (x^{2} + x) + C

\frac{d}{d x} (2 - y - 2 x)

x \to - 2 lim (x (- 2))

f (x) = (\frac{x + 4}{x - 4})^{5}

(1 - x^{2}) y^{^{'}} + x y = a x

What is the integral of 1/4 (2x+1) ?
The integral of 1/4 (2x+1) is 1/4 (x^2+x)+C