What is the integral of (14x+9)/((2x+1)^2) ?

The integral of (14x+9)/((2x+1)^2) is 7/2 ln|2x+1|-1/(2x+1)+C

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integral of (14x+9)/((2x+1)^2)

\int \frac{14 x + 9}{( 2 x + 1 ) ^{2}} d x

\frac{7}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2 x + 1} + C

Solution steps

\int \frac{14 x + 9}{( 2 x + 1 ) ^{2}} d x

Expand

\frac{14 x + 9}{( 2 x + 1 ) ^{2}} : \frac{14 x}{( 2 x + 1 ) ^{2}} + \frac{9}{( 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{14 x}{( 2 x + 1 ) ^{2}} d x + \int \frac{9}{( 2 x + 1 ) ^{2}} d x

\int \frac{14 x}{( 2 x + 1 ) ^{2}} d x = \frac{7}{2} (ln ∣ 2 x + 1 ∣ + \frac{1}{2 x + 1})

\int \frac{9}{( 2 x + 1 ) ^{2}} d x = - \frac{9}{2 ( 2 x + 1 )}

= \frac{7}{2} (ln ∣ 2 x + 1 ∣ + \frac{1}{2 x + 1}) - \frac{9}{2 ( 2 x + 1 )}

Simplify

\frac{7}{2} (ln ∣ 2 x + 1 ∣ + \frac{1}{2 x + 1}) - \frac{9}{2 ( 2 x + 1 )} : \frac{7}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2 x + 1}

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

Add a constant to the solution

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

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What is the integral of (14x+9)/((2x+1)^2) ?
The integral of (14x+9)/((2x+1)^2) is 7/2 ln|2x+1|-1/(2x+1)+C