What is the integral of sinh(1+4x) ?

The integral of sinh(1+4x) is 1/4 cosh(1+4x)+C

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integral of sinh(1+4x)

\int sinh (1 + 4 x) d x

\frac{1}{4} cosh (1 + 4 x) + C

Solution steps

\int sinh (1 + 4 x) d x

Apply u-substitution

: \int sinh (u) \frac{1}{4} d u

= \int sinh (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sinh (u) d u

Use the common integral:

\int sinh (u) d u = cosh (u)

= \frac{1}{4} cosh (u)

Substitute back

u = 1 + 4 x

= \frac{1}{4} cosh (1 + 4 x)

Add a constant to the solution

= \frac{1}{4} cosh (1 + 4 x) + C

\frac{d}{d x} (a (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2}))

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

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

(t^{2} + \frac{1}{2})^{2}

(x^{2} + 2 y^{2}) d x - x y d y = 0

What is the integral of sinh(1+4x) ?
The integral of sinh(1+4x) is 1/4 cosh(1+4x)+C