What is the integral of ((x^2+1))/((x-5)(x-4)^2) ?

The integral of ((x^2+1))/((x-5)(x-4)^2) is 26ln|x-5|-25ln|x-4|+(17)/(x-4)+C

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integral of ((x^2+1))/((x-5)(x-4)^2)

\int \frac{( x ^{2} + 1 )}{( x - 5 ) ( x - 4 ) ^{2}} d x

26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ + \frac{17}{x - 4} + C

Solution steps

\int \frac{x ^{2} + 1}{( x - 5 ) ( x - 4 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 1}{( x - 5 ) ( x - 4 ) ^{2}} : \frac{26}{x - 5} - \frac{25}{x - 4} - \frac{17}{( x - 4 ) ^{2}}

= \int \frac{26}{x - 5} - \frac{25}{x - 4} - \frac{17}{( x - 4 ) ^{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{26}{x - 5} d x - \int \frac{25}{x - 4} d x - \int \frac{17}{( x - 4 ) ^{2}} d x

\int \frac{26}{x - 5} d x = 26 ln ∣ x - 5 ∣

\int \frac{25}{x - 4} d x = 25 ln ∣ x - 4 ∣

\int \frac{17}{( x - 4 ) ^{2}} d x = - \frac{17}{x - 4}

= 26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ - (- \frac{17}{x - 4})

Simplify

= 26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ + \frac{17}{x - 4}

Add a constant to the solution

= 26 ln ∣ x - 5 ∣ - 25 ln ∣ x - 4 ∣ + \frac{17}{x - 4} + C

\frac{d}{d x} (- \frac{1}{2 ( 1 + x ) ^{\frac{3}{2}}})

f (x) = - \frac{6}{( 1 + 3 x ) ^{2}}

\frac{d}{d x} (ln (\frac{( 2 )}{3 - x ^{2}}))

\frac{3}{t ^{3}}

What is the integral of ((x^2+1))/((x-5)(x-4)^2) ?
The integral of ((x^2+1))/((x-5)(x-4)^2) is 26ln|x-5|-25ln|x-4|+(17)/(x-4)+C