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

The integral of (x^2-4)/(4x) is (x^2)/8-ln|x|+C

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integral of (x^2-4)/(4x)

\int \frac{x ^{2} - 4}{4 x} d x

\frac{x ^{2}}{8} - ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 4}{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{x ^{2} - 4}{x} d x

Expand

\frac{x ^{2} - 4}{x} : x - \frac{4}{x}

= \frac{1}{4} \cdot \int x - \frac{4}{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 x d x - \int \frac{4}{x} d x)

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

Simplify

\frac{1}{4} (\frac{x ^{2}}{2} - 4 ln ∣ x ∣) : \frac{x ^{2}}{8} - ln ∣ x ∣

= \frac{x ^{2}}{8} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{8} - ln ∣ x ∣ + C

n = 0 \sum \infty \frac{n + 1}{2 n - 3}

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

\frac{d x}{d t} = 7 (x^{2} + 1), x (\frac{π}{4}) = 1

6 t^{2} - 4 t + 9

f (t) = 2 e^{- t} cos (4 t)

What is the integral of (x^2-4)/(4x) ?
The integral of (x^2-4)/(4x) is (x^2)/8-ln|x|+C