What is the integral of (x^2+1)/((x-7)(x-6)) ?

The integral of (x^2+1)/((x-7)(x-6)) is x+50ln|x-7|-37ln|x-6|+C

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integral of (x^2+1)/((x-7)(x-6))

\int \frac{x ^{2} + 1}{( x - 7 ) ( x - 6 )} d x

x + 50 ln ∣ x - 7 ∣ - 37 ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{x ^{2} + 1}{( x - 7 ) ( x - 6 )} d x

Take the partial fraction of

\frac{x ^{2} + 1}{( x - 7 ) ( x - 6 )} : 1 + \frac{50}{x - 7} - \frac{37}{x - 6}

= \int 1 + \frac{50}{x - 7} - \frac{37}{x - 6} 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 1 d x + \int \frac{50}{x - 7} d x - \int \frac{37}{x - 6} d x

\int 1 d x = x

\int \frac{50}{x - 7} d x = 50 ln ∣ x - 7 ∣

\int \frac{37}{x - 6} d x = 37 ln ∣ x - 6 ∣

= x + 50 ln ∣ x - 7 ∣ - 37 ln ∣ x - 6 ∣

Add a constant to the solution

= x + 50 ln ∣ x - 7 ∣ - 37 ln ∣ x - 6 ∣ + C

What is the integral of (x^2+1)/((x-7)(x-6)) ?
The integral of (x^2+1)/((x-7)(x-6)) is x+50ln|x-7|-37ln|x-6|+C