What is the integral of (10)/((x-1)(x^2+9)) ?

The integral of (10)/((x-1)(x^2+9)) is ln|x-1|-1/2 ln|x^2+9|-1/3 arctan(x/3)+C

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integral of (10)/((x-1)(x^2+9))

\int \frac{10}{( x - 1 ) ( x ^{2} + 9 )} d x

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

Solution steps

\int \frac{10}{( x - 1 ) ( x ^{2} + 9 )} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{( x - 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x ^{2} + 9 )} : \frac{1}{10 ( x - 1 )} + \frac{- x - 1}{10 ( x ^{2} + 9 )}

= 10 \cdot \int \frac{1}{10 ( x - 1 )} + \frac{- x - 1}{10 ( 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

= 10 (\int \frac{1}{10 ( x - 1 )} d x + \int \frac{- x - 1}{10 ( x ^{2} + 9 )} d x)

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{- x - 1}{10 ( x ^{2} + 9 )} d x = \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

= 10 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})))

Simplify

10 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

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

Add a constant to the solution

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

cos (π) (x), 0

\frac{d}{d x} (120 x^{3} - 240 x^{2} + 120 x)

\int 36 x d x

\int cos (x) + \frac{1}{5} x d x

7 x + 1

What is the integral of (10)/((x-1)(x^2+9)) ?
The integral of (10)/((x-1)(x^2+9)) is ln|x-1|-1/2 ln|x^2+9|-1/3 arctan(x/3)+C