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

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

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

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

ln x^{2} + 4 - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x ^{2} - 4}{x ( x ^{2} + 4 )} : \frac{x ^{2}}{x ( x ^{2} + 4 )} - \frac{4}{x ( x ^{2} + 4 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

= ln x^{2} + 4 - ln ∣ x ∣

Add a constant to the solution

= ln x^{2} + 4 - ln ∣ x ∣ + C

\int cos^{5} (θ) sin (θ) d θ

y = x, y = 2 x 49 - x^{2}

y^{^{''}} + 25 y = 0, y (π) = 3, y^{^{'}} (π) = - 3

\int_{0}^{4} (- x^{2} + 4 x) d x

(1) (- 1.5)

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