What is the integral of e^{4x-7} ?

The integral of e^{4x-7} is 1/4 e^{4x-7}+C

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integral of e^{4x-7}

\int e^{4 x - 7} d x

\frac{1}{4} e^{4 x - 7} + C

Solution steps

\int e^{4 x - 7} 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 - 7

= \frac{1}{4} e^{4 x - 7}

Add a constant to the solution

= \frac{1}{4} e^{4 x - 7} + C

f (x) = cot^{2} (x)

\int e^{0.04 x} d x

y^{^{'}} + y = 2 e^{- t}

\int_{0}^{2} t e^{- 1} d t

f (x) = 5 x + x^{2}