What is the integral of e^{4x-5} ?

The integral of e^{4x-5} is 1/4 e^{4x-5}+C

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integral of e^{4x-5}

\int e^{4 x - 5} d x

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

Solution steps

\int e^{4 x - 5} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= \int e^{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 e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{4} e^{u}

Substitute back

u = 4 x - 5

= \frac{1}{4} e^{4 x - 5}

Add a constant to the solution

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

\int \frac{x ^{2} - 8}{4 x - x ^{3}} d x

- \frac{x}{6 y}

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 1)^{5})

x y^{^{'}} + 2 y = - 4 cos (x)

\int arcsin (\frac{2 x}{5}) d x