What is the integral of (25)/((x-4)(x^2+9)) ?

The integral of (25)/((x-4)(x^2+9)) is ln|x-4|-1/2 ln|x^2+9|-4/3 arctan(x/3)+C

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integral of (25)/((x-4)(x^2+9))

\int \frac{25}{( x - 4 ) ( x ^{2} + 9 )} d x

ln ∣ x - 4 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{25}{( x - 4 ) ( x ^{2} + 9 )} d x

Take the constant out:

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

= 25 \cdot \int \frac{1}{( x - 4 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{1}{( x - 4 ) ( x ^{2} + 9 )} : \frac{1}{25 ( x - 4 )} + \frac{- x - 4}{25 ( x ^{2} + 9 )}

= 25 \cdot \int \frac{1}{25 ( x - 4 )} + \frac{- x - 4}{25 ( x ^{2} + 9 )} 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

= 25 (\int \frac{1}{25 ( x - 4 )} d x + \int \frac{- x - 4}{25 ( x ^{2} + 9 )} d x)

\int \frac{1}{25 ( x - 4 )} d x = \frac{1}{25} ln ∣ x - 4 ∣

\int \frac{- x - 4}{25 ( x ^{2} + 9 )} d x = \frac{1}{25} (- \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3}))

= 25 (\frac{1}{25} ln ∣ x - 4 ∣ + \frac{1}{25} (- \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3})))

Simplify

25 (\frac{1}{25} ln ∣ x - 4 ∣ + \frac{1}{25} (- \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3}))) : ln ∣ x - 4 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3})

= ln ∣ x - 4 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3})

Add a constant to the solution

= ln ∣ x - 4 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3}) + C

f (x) = x + 40

n \to \infty lim (1 + \frac{1}{n})

y^{^{'}} + 2 t y = 2 t

y = 7 + 5 x^{2} - 2 x^{3}

x, x^{3}

What is the integral of (25)/((x-4)(x^2+9)) ?
The integral of (25)/((x-4)(x^2+9)) is ln|x-4|-1/2 ln|x^2+9|-4/3 arctan(x/3)+C