What is the integral of-(3x-1)/((x-5)(x-7)^2) ?

The integral of-(3x-1)/((x-5)(x-7)^2) is -7/2 ln|x-5|+7/2 ln|x-7|+(10)/(x-7)+C

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integral of-(3x-1)/((x-5)(x-7)^2)

\int - \frac{3 x - 1}{( x - 5 ) ( x - 7 ) ^{2}} d x

- \frac{7}{2} ln ∣ x - 5 ∣ + \frac{7}{2} ln ∣ x - 7 ∣ + \frac{10}{x - 7} + C

Solution steps

\int - \frac{3 x - 1}{( x - 5 ) ( x - 7 ) ^{2}} d x

Take the constant out:

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

= - \int \frac{3 x - 1}{( x - 5 ) ( x - 7 ) ^{2}} d x

Take the partial fraction of

\frac{3 x - 1}{( x - 5 ) ( x - 7 ) ^{2}} : \frac{7}{2 ( x - 5 )} - \frac{7}{2 ( x - 7 )} + \frac{10}{( x - 7 ) ^{2}}

= - \int \frac{7}{2 ( x - 5 )} - \frac{7}{2 ( x - 7 )} + \frac{10}{( x - 7 ) ^{2}} 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

= - (\int \frac{7}{2 ( x - 5 )} d x - \int \frac{7}{2 ( x - 7 )} d x + \int \frac{10}{( x - 7 ) ^{2}} d x)

\int \frac{7}{2 ( x - 5 )} d x = \frac{7}{2} ln ∣ x - 5 ∣

\int \frac{7}{2 ( x - 7 )} d x = \frac{7}{2} ln ∣ x - 7 ∣

\int \frac{10}{( x - 7 ) ^{2}} d x = - \frac{10}{x - 7}

= - (\frac{7}{2} ln ∣ x - 5 ∣ - \frac{7}{2} ln ∣ x - 7 ∣ - \frac{10}{x - 7})

Simplify

= - \frac{7}{2} ln ∣ x - 5 ∣ + \frac{7}{2} ln ∣ x - 7 ∣ + \frac{10}{x - 7}

Add a constant to the solution

= - \frac{7}{2} ln ∣ x - 5 ∣ + \frac{7}{2} ln ∣ x - 7 ∣ + \frac{10}{x - 7} + C

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What is the integral of-(3x-1)/((x-5)(x-7)^2) ?
The integral of-(3x-1)/((x-5)(x-7)^2) is -7/2 ln|x-5|+7/2 ln|x-7|+(10)/(x-7)+C