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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\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}))

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

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

Simplify

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

Add a constant to the solution

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

g (x) = tan (2 x) \cdot sin (2 x)

(2 y^{2} x - y) d x + x d y = 0

y = 2 x - x^{2}

\frac{\partial}{\partial x} (\frac{10000 x}{y ^{2}})

n = 1 \sum \infty (\frac{8}{5})^{n - 1}

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