What is the integral of (5x^2)/((x-2)(x^2+4)) ?

The integral of (5x^2)/((x-2)(x^2+4)) is 5(1/2 ln|x-2|+1/2 (1/2 ln|(x^2)/4+1|+arctan(x/2)))+C

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integral of (5x^2)/((x-2)(x^2+4))

\int \frac{5 x ^{2}}{( x - 2 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{5 x ^{2}}{( x - 2 ) ( x ^{2} + 4 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{( x - 2 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

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

= 5 \cdot \int \frac{1}{2 ( x - 2 )} + \frac{x + 2}{2 ( 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

= 5 (\int \frac{1}{2 ( x - 2 )} d x + \int \frac{x + 2}{2 ( x ^{2} + 4 )} d x)

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

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

= 5 (\frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2})))

Add a constant to the solution

= 5 (\frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2}))) + C

\frac{d}{d x} (x + 1 \cdot e^{x})

\frac{7}{x ^{2}}

\int tan^{5} (4 x) d x

\int \frac{x}{3 x ^{2} + 8 x - 3} d x

x^{^{''}} + 3 x^{^{'}} + 2 x = 0

What is the integral of (5x^2)/((x-2)(x^2+4)) ?
The integral of (5x^2)/((x-2)(x^2+4)) is 5(1/2 ln|x-2|+1/2 (1/2 ln|(x^2)/4+1|+arctan(x/2)))+C