What is the integral of-8e^{x+4}-5/(x+7)-5sec^2(x+6) ?

The integral of-8e^{x+4}-5/(x+7)-5sec^2(x+6) is -8e^{4+x}-5ln|x+7|-5(-6+tan(x+6))+C

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integral of-8e^{x+4}-5/(x+7)-5sec^2(x+6)

\int - 8 e^{x + 4} - \frac{5}{x + 7} - 5 sec^{2} (x + 6) d x

- 8 e^{4 + x} - 5 ln ∣ x + 7 ∣ - 5 (- 6 + tan (x + 6)) + C

Solution steps

\int - 8 e^{x + 4} - \frac{5}{x + 7} - 5 sec^{2} (x + 6) d x

Rewrite using trig identities

= \int - 8 e^{x + 4} - \frac{5}{x + 7} - 5 (1 + tan^{2} (x + 6)) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 8 e^{x + 4} d x - \int \frac{5}{x + 7} d x - \int 5 (1 + tan^{2} (x + 6)) d x

\int 8 e^{x + 4} d x = 8 e^{4 + x}

\int \frac{5}{x + 7} d x = 5 ln ∣ x + 7 ∣

\int 5 (1 + tan^{2} (x + 6)) d x = 5 (- 6 + tan (x + 6))

= - 8 e^{4 + x} - 5 ln ∣ x + 7 ∣ - 5 (- 6 + tan (x + 6))

Add a constant to the solution

= - 8 e^{4 + x} - 5 ln ∣ x + 7 ∣ - 5 (- 6 + tan (x + 6)) + C

x \to - 1 + lim (x + 4)

\frac{d x}{d y} = x y sin (y^{2})

\frac{d}{d x} (\frac{2 e ^{- \frac{1}{x ^{2}}}}{x ^{3}})

\frac{\partial}{\partial z} (4 z)

\int - tan (y) d y

What is the integral of-8e^{x+4}-5/(x+7)-5sec^2(x+6) ?
The integral of-8e^{x+4}-5/(x+7)-5sec^2(x+6) is -8e^{4+x}-5ln|x+7|-5(-6+tan(x+6))+C