What is the integral of (13)/((x-3)(x^2+4)) ?

The integral of (13)/((x-3)(x^2+4)) is ln|x-3|-1/2 ln|x^2+4|-3/2 arctan(x/2)+C

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integral of (13)/((x-3)(x^2+4))

\int \frac{13}{( x - 3 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{13}{( x - 3 ) ( x ^{2} + 4 )} d x

Take the constant out:

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

= 13 \cdot \int \frac{1}{( x - 3 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{1}{( x - 3 ) ( x ^{2} + 4 )} : \frac{1}{13 ( x - 3 )} + \frac{- x - 3}{13 ( x ^{2} + 4 )}

= 13 \cdot \int \frac{1}{13 ( x - 3 )} + \frac{- x - 3}{13 ( 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

= 13 (\int \frac{1}{13 ( x - 3 )} d x + \int \frac{- x - 3}{13 ( x ^{2} + 4 )} d x)

\int \frac{1}{13 ( x - 3 )} d x = \frac{1}{13} ln ∣ x - 3 ∣

\int \frac{- x - 3}{13 ( x ^{2} + 4 )} d x = \frac{1}{13} (- \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}))

= 13 (\frac{1}{13} ln ∣ x - 3 ∣ + \frac{1}{13} (- \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2})))

Simplify

13 (\frac{1}{13} ln ∣ x - 3 ∣ + \frac{1}{13} (- \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}))) : ln ∣ x - 3 ∣ - \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2})

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

Add a constant to the solution

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

f (x) = \frac{2 ln ( x )}{x}

\int e^{18 x} - 4 d x

x \to 2 lim (6 \cdot a + x)

\frac{( 18 )}{s ( s ^{2} + 6 s + 18 )}

3 x y^{^{'}} + y = 12 x

What is the integral of (13)/((x-3)(x^2+4)) ?
The integral of (13)/((x-3)(x^2+4)) is ln|x-3|-1/2 ln|x^2+4|-3/2 arctan(x/2)+C