What is the integral of (10)/((4x-4)(x^2+9)) ?

The integral of (10)/((4x-4)(x^2+9)) is 10(1/40 ln|x-1|+1/40 (-1/2 ln|x^2+9|-1/3 arctan(x/3)))+C

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integral of (10)/((4x-4)(x^2+9))

\int \frac{10}{( 4 x - 4 ) ( x ^{2} + 9 )} d x

10 (\frac{1}{40} ln ∣ x - 1 ∣ + \frac{1}{40} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) + C

Solution steps

\int \frac{10}{( 4 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

= 10 \cdot \int \frac{1}{( 4 x - 4 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{1}{( 4 x - 4 ) ( x ^{2} + 9 )} : \frac{1}{40 ( x - 1 )} + \frac{- x - 1}{40 ( x ^{2} + 9 )}

= 10 \cdot \int \frac{1}{40 ( x - 1 )} + \frac{- x - 1}{40 ( 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}{40 ( x - 1 )} d x + \int \frac{- x - 1}{40 ( x ^{2} + 9 )} d x)

\int \frac{1}{40 ( x - 1 )} d x = \frac{1}{40} ln ∣ x - 1 ∣

\int \frac{- x - 1}{40 ( x ^{2} + 9 )} d x = \frac{1}{40} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

= 10 (\frac{1}{40} ln ∣ x - 1 ∣ + \frac{1}{40} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})))

Add a constant to the solution

= 10 (\frac{1}{40} ln ∣ x - 1 ∣ + \frac{1}{40} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) + C

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What is the integral of (10)/((4x-4)(x^2+9)) ?
The integral of (10)/((4x-4)(x^2+9)) is 10(1/40 ln|x-1|+1/40 (-1/2 ln|x^2+9|-1/3 arctan(x/3)))+C