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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

y^{^{''}} - 2 y^{^{'}} + y = t e^{t}, y (0) = 1, y^{^{'}} (0) = 1

\int \frac{20}{x} d x

x \to 0 lim (5 + 2 e^{\frac{- 5}{x}})

\int cos^{2} (w x) d x

f (x) = 3 sec (x), at x = \frac{π}{6}

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