What is the integral of cosh(4x-1) ?

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

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integral of cosh(4x-1)

\int cosh (4 x - 1) d x

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

Solution steps

\int cosh (4 x - 1) d x

Rewrite using trig identities

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

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

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

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

Add a constant to the solution

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

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