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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

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\int 3 e^{- 3 x} x^{2} d x

\int 3 x^{2} ln (x + 2) d x

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