What is the integral of ((x^2+1))/((x-4)(x-3)^2) ?

The integral of ((x^2+1))/((x-4)(x-3)^2) is 17ln|x-4|-16ln|x-3|+(10)/(x-3)+C

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integral of ((x^2+1))/((x-4)(x-3)^2)

\int \frac{( x ^{2} + 1 )}{( x - 4 ) ( x - 3 ) ^{2}} d x

17 ln ∣ x - 4 ∣ - 16 ln ∣ x - 3 ∣ + \frac{10}{x - 3} + C

Solution steps

\int \frac{x ^{2} + 1}{( x - 4 ) ( x - 3 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 1}{( x - 4 ) ( x - 3 ) ^{2}} : \frac{17}{x - 4} - \frac{16}{x - 3} - \frac{10}{( x - 3 ) ^{2}}

= \int \frac{17}{x - 4} - \frac{16}{x - 3} - \frac{10}{( x - 3 ) ^{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{17}{x - 4} d x - \int \frac{16}{x - 3} d x - \int \frac{10}{( x - 3 ) ^{2}} d x

\int \frac{17}{x - 4} d x = 17 ln ∣ x - 4 ∣

\int \frac{16}{x - 3} d x = 16 ln ∣ x - 3 ∣

\int \frac{10}{( x - 3 ) ^{2}} d x = - \frac{10}{x - 3}

= 17 ln ∣ x - 4 ∣ - 16 ln ∣ x - 3 ∣ - (- \frac{10}{x - 3})

Simplify

= 17 ln ∣ x - 4 ∣ - 16 ln ∣ x - 3 ∣ + \frac{10}{x - 3}

Add a constant to the solution

= 17 ln ∣ x - 4 ∣ - 16 ln ∣ x - 3 ∣ + \frac{10}{x - 3} + C

y^{^{''}} - 10 y^{^{'}} + 34 y = 0, y (0) = - 2, y^{^{'}} (0) = - 3

\frac{3}{x}, \frac{3}{x ^{2}}, 6

y = 5, y = x^{2} + 1

(x - 2)^{2} - 3 x

What is the integral of ((x^2+1))/((x-4)(x-3)^2) ?
The integral of ((x^2+1))/((x-4)(x-3)^2) is 17ln|x-4|-16ln|x-3|+(10)/(x-3)+C