What is the integral of (3x^2+4x-6)/(x^2+16x+64) ?

The integral of (3x^2+4x-6)/(x^2+16x+64) is 3x-44ln|x+8|-(154)/(x+8)+24+C

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integral of (3x^2+4x-6)/(x^2+16x+64)

\int \frac{3 x ^{2} + 4 x - 6}{x ^{2} + 16 x + 64} d x

3 x - 44 ln ∣ x + 8 ∣ - \frac{154}{x + 8} + 24 + C

Solution steps

\int \frac{3 x ^{2} + 4 x - 6}{x ^{2} + 16 x + 64} d x

Expand

\frac{3 x ^{2} + 4 x - 6}{x ^{2} + 16 x + 64} : \frac{3 x ^{2}}{x ^{2} + 16 x + 64} + \frac{4 x}{x ^{2} + 16 x + 64} - \frac{6}{x ^{2} + 16 x + 64}

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 ^{2}}{x ^{2} + 16 x + 64} d x + \int \frac{4 x}{x ^{2} + 16 x + 64} d x - \int \frac{6}{x ^{2} + 16 x + 64} d x

\int \frac{3 x ^{2}}{x ^{2} + 16 x + 64} d x = 3 (x + 8 - 16 ln ∣ x + 8 ∣ - \frac{64}{x + 8})

\int \frac{4 x}{x ^{2} + 16 x + 64} d x = 4 (ln ∣ x + 8 ∣ + \frac{8}{x + 8})

\int \frac{6}{x ^{2} + 16 x + 64} d x = - \frac{6}{x + 8}

= 3 (x + 8 - 16 ln ∣ x + 8 ∣ - \frac{64}{x + 8}) + 4 (ln ∣ x + 8 ∣ + \frac{8}{x + 8}) - (- \frac{6}{x + 8})

Simplify

3 (x + 8 - 16 ln ∣ x + 8 ∣ - \frac{64}{x + 8}) + 4 (ln ∣ x + 8 ∣ + \frac{8}{x + 8}) - (- \frac{6}{x + 8}) : 3 x - 44 ln ∣ x + 8 ∣ - \frac{154}{x + 8} + 24

= 3 x - 44 ln ∣ x + 8 ∣ - \frac{154}{x + 8} + 24

Add a constant to the solution

= 3 x - 44 ln ∣ x + 8 ∣ - \frac{154}{x + 8} + 24 + C

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

\int \frac{x + 2}{x - 4} d x

\frac{d}{d x} (cos (e^{2 x}))

9 x - 8 x

(\frac{ln ^{2} ( x )}{x})^{^{'}}

What is the integral of (3x^2+4x-6)/(x^2+16x+64) ?
The integral of (3x^2+4x-6)/(x^2+16x+64) is 3x-44ln|x+8|-(154)/(x+8)+24+C