What is the integral of (-10x+2)/((-2x+1)^2) ?

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

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integral of (-10x+2)/((-2x+1)^2)

\int \frac{- 10 x + 2}{( - 2 x + 1 ) ^{2}} d x

- \frac{5}{2} (ln ∣ - 2 x + 1 ∣ + \frac{1}{- 2 x + 1}) + \frac{1}{- 2 x + 1} + C

Solution steps

\int \frac{- 10 x + 2}{( - 2 x + 1 ) ^{2}} d x

Expand

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

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

\int \frac{10 x}{( - 2 x + 1 ) ^{2}} d x = \frac{5}{2} (ln ∣ - 2 x + 1 ∣ + \frac{1}{- 2 x + 1})

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

= - \frac{5}{2} (ln ∣ - 2 x + 1 ∣ + \frac{1}{- 2 x + 1}) + \frac{1}{- 2 x + 1}

Add a constant to the solution

= - \frac{5}{2} (ln ∣ - 2 x + 1 ∣ + \frac{1}{- 2 x + 1}) + \frac{1}{- 2 x + 1} + C

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