What is the integral of 10sinh(4x-ln(2)) ?

The integral of 10sinh(4x-ln(2)) is 5/2 (e^{-4x}+1/4 e^{4x})+C

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integral of 10sinh(4x-ln(2))

\int 10 sinh (4 x - ln (2)) d x

\frac{5}{2} (e^{- 4 x} + \frac{1}{4} e^{4 x}) + C

Solution steps

\int 10 sinh (4 x - ln (2)) d x

Take the constant out:

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

= 10 \cdot \int sinh (4 x - ln (2)) d x

Rewrite using trig identities

= 10 \cdot \int \frac{1}{4} (- 4 e^{- 4 x} + e^{4 x}) d x

Take the constant out:

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

= 10 \cdot \frac{1}{4} \cdot \int - 4 e^{- 4 x} + e^{4 x} 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

= 10 \cdot \frac{1}{4} (- \int 4 e^{- 4 x} d x + \int e^{4 x} d x)

\int 4 e^{- 4 x} d x = - e^{- 4 x}

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

= 10 \cdot \frac{1}{4} (- (- e^{- 4 x}) + \frac{1}{4} e^{4 x})

Simplify

10 \cdot \frac{1}{4} (- (- e^{- 4 x}) + \frac{1}{4} e^{4 x}) : \frac{5}{2} (e^{- 4 x} + \frac{1}{4} e^{4 x})

= \frac{5}{2} (e^{- 4 x} + \frac{1}{4} e^{4 x})

Add a constant to the solution

= \frac{5}{2} (e^{- 4 x} + \frac{1}{4} e^{4 x}) + C

\int ln (x) x d x

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

\frac{\partial}{\partial x} (x^{2} + 2 x y + y^{4})

f (x) = \frac{4}{x}, (- 9, - \frac{4}{9})

\frac{\partial}{\partial y} (x + y - 8)

What is the integral of 10sinh(4x-ln(2)) ?
The integral of 10sinh(4x-ln(2)) is 5/2 (e^{-4x}+1/4 e^{4x})+C