What is the integral of (3x+3)/(x^2+8x+16) ?

The integral of (3x+3)/(x^2+8x+16) is 3ln|x+4|+9/(x+4)+C

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integral of (3x+3)/(x^2+8x+16)

\int \frac{3 x + 3}{x ^{2} + 8 x + 16} d x

3 ln ∣ x + 4 ∣ + \frac{9}{x + 4} + C

Solution steps

\int \frac{3 x + 3}{x ^{2} + 8 x + 16} d x

Expand

\frac{3 x + 3}{x ^{2} + 8 x + 16} : \frac{3 x}{x ^{2} + 8 x + 16} + \frac{3}{x ^{2} + 8 x + 16}

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{3 x}{x ^{2} + 8 x + 16} d x + \int \frac{3}{x ^{2} + 8 x + 16} d x

\int \frac{3 x}{x ^{2} + 8 x + 16} d x = 3 (ln ∣ x + 4 ∣ + \frac{4}{x + 4})

\int \frac{3}{x ^{2} + 8 x + 16} d x = - \frac{3}{x + 4}

= 3 (ln ∣ x + 4 ∣ + \frac{4}{x + 4}) - \frac{3}{x + 4}

Simplify

3 (ln ∣ x + 4 ∣ + \frac{4}{x + 4}) - \frac{3}{x + 4} : 3 ln ∣ x + 4 ∣ + \frac{9}{x + 4}

= 3 ln ∣ x + 4 ∣ + \frac{9}{x + 4}

Add a constant to the solution

= 3 ln ∣ x + 4 ∣ + \frac{9}{x + 4} + C

\int sec^{2} (π x) + e^{- 3 x} d x

\int sec^{2} (2 a) x d x

\frac{1}{u}

\frac{1}{t}

\int \frac{1}{- x ^{2} - 8 x} d x

What is the integral of (3x+3)/(x^2+8x+16) ?
The integral of (3x+3)/(x^2+8x+16) is 3ln|x+4|+9/(x+4)+C