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

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

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

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

- 5 ln ∣ x + 1 ∣ - \frac{7}{x + 1} + C

Solution steps

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

Expand

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

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

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

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

Simplify

- 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{2}{x + 1} : - 5 ln ∣ x + 1 ∣ - \frac{7}{x + 1}

= - 5 ln ∣ x + 1 ∣ - \frac{7}{x + 1}

Add a constant to the solution

= - 5 ln ∣ x + 1 ∣ - \frac{7}{x + 1} + C

\frac{d}{d x} (cos (x - \frac{π}{6}))

y^{^{'}} = y e^{x}

y = (1 - 5 x)^{6}

y = 7 x + 1, (5, 6)

x \to 0 lim ((1 + 3 x)^{\frac{4}{x}})

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