What is the integral of e^{7x-1} ?

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

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integral of e^{7x-1}

\int e^{7 x - 1} d x

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

Solution steps

\int e^{7 x - 1} d x

Apply u-substitution

: \int e^{u} \frac{1}{7} d u

= \int e^{u} \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{7} e^{u}

Substitute back

u = 7 x - 1

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

Add a constant to the solution

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

\int ∣ 1 - x ∣ d x

\frac{d}{d x} (\frac{\frac{d}{d x} ( 3 )}{x})

\frac{d}{d x} (\frac{1}{3 + 2 x})

y^{^{'}} = - x y^{3}

x \to 0 - lim (g (x))