What is the integral of 6/((x-7)(x-1)) ?

The integral of 6/((x-7)(x-1)) is ln|x-7|-ln|x-1|+C

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integral of 6/((x-7)(x-1))

\int \frac{6}{( x - 7 ) ( x - 1 )} d x

ln ∣ x - 7 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{6}{( x - 7 ) ( x - 1 )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( x - 7 ) ( x - 1 )} d x

Take the partial fraction of

\frac{1}{( x - 7 ) ( x - 1 )} : \frac{1}{6 ( x - 7 )} - \frac{1}{6 ( x - 1 )}

= 6 \cdot \int \frac{1}{6 ( x - 7 )} - \frac{1}{6 ( 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

= 6 (\int \frac{1}{6 ( x - 7 )} d x - \int \frac{1}{6 ( x - 1 )} d x)

\int \frac{1}{6 ( x - 7 )} d x = \frac{1}{6} ln ∣ x - 7 ∣

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

= 6 (\frac{1}{6} ln ∣ x - 7 ∣ - \frac{1}{6} ln ∣ x - 1 ∣)

Simplify

6 (\frac{1}{6} ln ∣ x - 7 ∣ - \frac{1}{6} ln ∣ x - 1 ∣) : ln ∣ x - 7 ∣ - ln ∣ x - 1 ∣

= ln ∣ x - 7 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x - 7 ∣ - ln ∣ x - 1 ∣ + C

\int x^{3} x^{2} + 47 d x

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\int \frac{4 x - 1}{2 x ^{2} - x + 1} d x

What is the integral of 6/((x-7)(x-1)) ?
The integral of 6/((x-7)(x-1)) is ln|x-7|-ln|x-1|+C