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

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

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integral of (4x^2+1)/(4x)

\int \frac{4 x ^{2} + 1}{4 x} d x

\frac{1}{4} (2 x^{2} + ln ∣ x ∣) + C

Solution steps

\int \frac{4 x ^{2} + 1}{4 x} 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 \frac{4 x ^{2} + 1}{x} d x

Expand

\frac{4 x ^{2} + 1}{x} : 4 x + \frac{1}{x}

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{4} (2 x^{2} + ln ∣ x ∣)

Add a constant to the solution

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

\int 3 x^{2} - 5 x + \frac{7}{x} + 2 e^{3} d x

\frac{\partial}{\partial x} (\frac{( x + y )}{x y})

x \to 2 + lim (x^{2} + 1)

y = 4 x - 3 x^{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

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