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

The integral of (x^2+1)/((x-3)(x-2)^2) is 10ln|x-3|-9ln|x-2|+5/(x-2)+C

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

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

10 ln ∣ x - 3 ∣ - 9 ln ∣ x - 2 ∣ + \frac{5}{x - 2} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 1}{( x - 3 ) ( x - 2 ) ^{2}} : \frac{10}{x - 3} - \frac{9}{x - 2} - \frac{5}{( x - 2 ) ^{2}}

= \int \frac{10}{x - 3} - \frac{9}{x - 2} - \frac{5}{( x - 2 ) ^{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{10}{x - 3} d x - \int \frac{9}{x - 2} d x - \int \frac{5}{( x - 2 ) ^{2}} d x

\int \frac{10}{x - 3} d x = 10 ln ∣ x - 3 ∣

\int \frac{9}{x - 2} d x = 9 ln ∣ x - 2 ∣

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

= 10 ln ∣ x - 3 ∣ - 9 ln ∣ x - 2 ∣ - (- \frac{5}{x - 2})

Simplify

= 10 ln ∣ x - 3 ∣ - 9 ln ∣ x - 2 ∣ + \frac{5}{x - 2}

Add a constant to the solution

= 10 ln ∣ x - 3 ∣ - 9 ln ∣ x - 2 ∣ + \frac{5}{x - 2} + C

f (x) = (- x^{2} + 9)^{2}

\int cosh (3 x) d x

x \frac{d y}{d x} - y = x^{2} - y^{2}

\int sin^{3} (x) cos^{\frac{3}{2}} (x) d x

x \frac{d y}{d x} - 3 y = x y^{4}

What is the integral of (x^2+1)/((x-3)(x-2)^2) ?
The integral of (x^2+1)/((x-3)(x-2)^2) is 10ln|x-3|-9ln|x-2|+5/(x-2)+C