What is the integral of (2x+1)/((x+7)^2) ?

The integral of (2x+1)/((x+7)^2) is 2ln|x+7|+(13)/(x+7)+C

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integral of (2x+1)/((x+7)^2)

\int \frac{2 x + 1}{( x + 7 ) ^{2}} d x

2 ln ∣ x + 7 ∣ + \frac{13}{x + 7} + C

Solution steps

\int \frac{2 x + 1}{( x + 7 ) ^{2}} d x

Expand

\frac{2 x + 1}{( x + 7 ) ^{2}} : \frac{2 x}{( x + 7 ) ^{2}} + \frac{1}{( x + 7 ) ^{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{2 x}{( x + 7 ) ^{2}} d x + \int \frac{1}{( x + 7 ) ^{2}} d x

\int \frac{2 x}{( x + 7 ) ^{2}} d x = 2 (ln ∣ x + 7 ∣ + \frac{7}{x + 7})

\int \frac{1}{( x + 7 ) ^{2}} d x = - \frac{1}{x + 7}

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

Simplify

2 (ln ∣ x + 7 ∣ + \frac{7}{x + 7}) - \frac{1}{x + 7} : 2 ln ∣ x + 7 ∣ + \frac{13}{x + 7}

= 2 ln ∣ x + 7 ∣ + \frac{13}{x + 7}

Add a constant to the solution

= 2 ln ∣ x + 7 ∣ + \frac{13}{x + 7} + C

\frac{d}{d x} (arcsin (x + y))

\frac{d y}{d x} = - \frac{1 + y}{1 + x + y}

4 y^{^{''}} + 9 y^{^{'}} = 0

\frac{\partial}{\partial x} (e^{1 x} cos (2 y))

y^{^{''}} + y = 6 sin (2 t) + 4 t cos (2 t)

What is the integral of (2x+1)/((x+7)^2) ?
The integral of (2x+1)/((x+7)^2) is 2ln|x+7|+(13)/(x+7)+C