What is the integral of (x^2-5x+16)/((2x+1)(x-2)^2) ?

The integral of (x^2-5x+16)/((2x+1)(x-2)^2) is 3/2 ln|2x+1|-ln|x-2|-2/(x-2)+C

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integral of (x^2-5x+16)/((2x+1)(x-2)^2)

\int \frac{x ^{2} - 5 x + 16}{( 2 x + 1 ) ( x - 2 ) ^{2}} d x

\frac{3}{2} ln ∣ 2 x + 1 ∣ - ln ∣ x - 2 ∣ - \frac{2}{x - 2} + C

Solution steps

\int \frac{x ^{2} - 5 x + 16}{( 2 x + 1 ) ( x - 2 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - 5 x + 16}{( 2 x + 1 ) ( x - 2 ) ^{2}} : \frac{3}{2 x + 1} - \frac{1}{x - 2} + \frac{2}{( x - 2 ) ^{2}}

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

\int \frac{3}{2 x + 1} d x = \frac{3}{2} ln ∣ 2 x + 1 ∣

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

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

= \frac{3}{2} ln ∣ 2 x + 1 ∣ - ln ∣ x - 2 ∣ - \frac{2}{x - 2}

Add a constant to the solution

= \frac{3}{2} ln ∣ 2 x + 1 ∣ - ln ∣ x - 2 ∣ - \frac{2}{x - 2} + C

\frac{\partial}{\partial y} (e^{x} y)

\frac{x d y}{d x} + y = \frac{1}{y ^{2}}

f (x) = \frac{x ^{2}}{x + 4}, at x = 2

\frac{d y}{d x} - 2 y = 4 - x

What is the integral of (x^2-5x+16)/((2x+1)(x-2)^2) ?
The integral of (x^2-5x+16)/((2x+1)(x-2)^2) is 3/2 ln|2x+1|-ln|x-2|-2/(x-2)+C