What is the integral of (7x+8)/(7x^2+16x+9) ?

The integral of (7x+8)/(7x^2+16x+9) is 1/2 ln|7x+7|+1/2 ln|7x+9|-1+C

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integral of (7x+8)/(7x^2+16x+9)

\int \frac{7 x + 8}{7 x ^{2} + 16 x + 9} d x

\frac{1}{2} ln ∣ 7 x + 7 ∣ + \frac{1}{2} ln ∣ 7 x + 9 ∣ - 1 + C

Solution steps

\int \frac{7 x + 8}{7 x ^{2} + 16 x + 9} d x

Expand

\frac{7 x + 8}{7 x ^{2} + 16 x + 9} : \frac{7 x}{7 x ^{2} + 16 x + 9} + \frac{8}{7 x ^{2} + 16 x + 9}

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{7 x}{7 x ^{2} + 16 x + 9} d x + \int \frac{8}{7 x ^{2} + 16 x + 9} d x

\int \frac{7 x}{7 x ^{2} + 16 x + 9} d x = - \frac{7}{2} ln ∣ 7 x + 7 ∣ + \frac{9}{2} ln ∣ 7 x + 9 ∣ - 1

\int \frac{8}{7 x ^{2} + 16 x + 9} d x = - 8 (\frac{1}{2} ln ∣ 7 x + 9 ∣ - \frac{1}{2} ln ∣ 7 x + 7 ∣)

= - \frac{7}{2} ln ∣ 7 x + 7 ∣ + \frac{9}{2} ln ∣ 7 x + 9 ∣ - 1 - 8 (\frac{1}{2} ln ∣ 7 x + 9 ∣ - \frac{1}{2} ln ∣ 7 x + 7 ∣)

Simplify

- \frac{7}{2} ln ∣ 7 x + 7 ∣ + \frac{9}{2} ln ∣ 7 x + 9 ∣ - 1 - 8 (\frac{1}{2} ln ∣ 7 x + 9 ∣ - \frac{1}{2} ln ∣ 7 x + 7 ∣) : \frac{1}{2} ln ∣ 7 x + 7 ∣ + \frac{1}{2} ln ∣ 7 x + 9 ∣ - 1

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{x ^{11}})

x \to 3 + lim (\frac{18 - 6 x}{3 - x})

\int x^{2} - 8 x + 12 d x

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

What is the integral of (7x+8)/(7x^2+16x+9) ?
The integral of (7x+8)/(7x^2+16x+9) is 1/2 ln|7x+7|+1/2 ln|7x+9|-1+C