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

The integral of (2x+1)/((x-2)(x-3)^2) is 5ln|x-2|-5ln|x-3|-7/(x-3)+C

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

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

5 ln ∣ x - 2 ∣ - 5 ln ∣ x - 3 ∣ - \frac{7}{x - 3} + C

Solution steps

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

Take the partial fraction of

\frac{2 x + 1}{( x - 2 ) ( x - 3 ) ^{2}} : \frac{5}{x - 2} - \frac{5}{x - 3} + \frac{7}{( x - 3 ) ^{2}}

= \int \frac{5}{x - 2} - \frac{5}{x - 3} + \frac{7}{( 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{5}{x - 2} d x - \int \frac{5}{x - 3} d x + \int \frac{7}{( x - 3 ) ^{2}} d x

\int \frac{5}{x - 2} d x = 5 ln ∣ x - 2 ∣

\int \frac{5}{x - 3} d x = 5 ln ∣ x - 3 ∣

\int \frac{7}{( x - 3 ) ^{2}} d x = - \frac{7}{x - 3}

= 5 ln ∣ x - 2 ∣ - 5 ln ∣ x - 3 ∣ - \frac{7}{x - 3}

Add a constant to the solution

= 5 ln ∣ x - 2 ∣ - 5 ln ∣ x - 3 ∣ - \frac{7}{x - 3} + C

\frac{d}{d x} (x^{5} - 3 x^{3} + x - 1)

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

y = (3 x^{3} + 2 x)^{\frac{1}{2}}

f (x) = e^{- 9}

\int_{0}^{0.08} 2000 x d x

What is the integral of (2x+1)/((x-2)(x-3)^2) ?
The integral of (2x+1)/((x-2)(x-3)^2) is 5ln|x-2|-5ln|x-3|-7/(x-3)+C