What is the integral of (65)/((x-1)(x^2+64)) ?

The integral of (65)/((x-1)(x^2+64)) is ln|x-1|-1/2 ln|x^2+64|-1/8 arctan(x/8)+C

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integral of (65)/((x-1)(x^2+64))

\int \frac{65}{( x - 1 ) ( x ^{2} + 64 )} d x

ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8}) + C

Solution steps

\int \frac{65}{( x - 1 ) ( x ^{2} + 64 )} d x

Take the constant out:

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

= 65 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 64 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} + 64 )} : \frac{1}{65 ( x - 1 )} + \frac{- x - 1}{65 ( x ^{2} + 64 )}

= 65 \cdot \int \frac{1}{65 ( x - 1 )} + \frac{- x - 1}{65 ( x ^{2} + 64 )} 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

= 65 (\int \frac{1}{65 ( x - 1 )} d x + \int \frac{- x - 1}{65 ( x ^{2} + 64 )} d x)

\int \frac{1}{65 ( x - 1 )} d x = \frac{1}{65} ln ∣ x - 1 ∣

\int \frac{- x - 1}{65 ( x ^{2} + 64 )} d x = \frac{1}{65} (- \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8}))

= 65 (\frac{1}{65} ln ∣ x - 1 ∣ + \frac{1}{65} (- \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8})))

Simplify

65 (\frac{1}{65} ln ∣ x - 1 ∣ + \frac{1}{65} (- \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8})

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8})

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 64 - \frac{1}{8} arctan (\frac{x}{8}) + C

x \to 25 lim (x - 25)

(x + 1) d y = (x + 6) d x

\frac{\partial}{\partial x} (x^{2} + x)

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 6

y = x^{2}, y = x^{4}

What is the integral of (65)/((x-1)(x^2+64)) ?
The integral of (65)/((x-1)(x^2+64)) is ln|x-1|-1/2 ln|x^2+64|-1/8 arctan(x/8)+C