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 t} (a t cos (4 t) + b t sin (4 t))

\frac{\partial}{\partial x} (- x^{2} sin (x^{2} y))

n \to \infty lim (\frac{4}{n})

\frac{d}{d x} ((2 x + e^{x}) (5 - x))

\frac{2 x ^{2}}{x ^{2} - 1}

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