What is the integral of sinh(4x+6) ?

The integral of sinh(4x+6) is 1/2 (1/4 e^{-4x-6}+1/4 e^{4x+6})+C

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integral of sinh(4x+6)

\int sinh (4 x + 6) d x

\frac{1}{2} (\frac{1}{4} e^{- 4 x - 6} + \frac{1}{4} e^{4 x + 6}) + C

Solution steps

\int sinh (4 x + 6) d x

Rewrite using trig identities

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int - e^{- 4 x - 6} + e^{4 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

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

\int e^{- 4 x - 6} d x = - \frac{1}{4} e^{- 4 x - 6}

\int e^{4 x + 6} d x = \frac{1}{4} e^{4 x + 6}

= \frac{1}{2} (- (- \frac{1}{4} e^{- 4 x - 6}) + \frac{1}{4} e^{4 x + 6})

Simplify

= \frac{1}{2} (\frac{1}{4} e^{- 4 x - 6} + \frac{1}{4} e^{4 x + 6})

Add a constant to the solution

= \frac{1}{2} (\frac{1}{4} e^{- 4 x - 6} + \frac{1}{4} e^{4 x + 6}) + C

\frac{d}{d x} (a sin (x) + b cos (x))

\int \frac{ln ( x ^{7} )}{x} d x

h \to 0 lim (h^{2} + h + 1)

y = - x^{2} + 8 x, y = x^{2} - 6 x

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

What is the integral of sinh(4x+6) ?
The integral of sinh(4x+6) is 1/2 (1/4 e^{-4x-6}+1/4 e^{4x+6})+C