What is the integral of e^{3x-5} ?

The integral of e^{3x-5} is 1/3 e^{3x-5}+C

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integral of e^{3x-5}

\int e^{3 x - 5} d x

\frac{1}{3} e^{3 x - 5} + C

Solution steps

\int e^{3 x - 5} d x

Apply u-substitution

= \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 3 x - 5

= \frac{1}{3} e^{3 x - 5}

Add a constant to the solution

= \frac{1}{3} e^{3 x - 5} + C

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} + 12 y = 0

\int \frac{1}{x ^{8} + x ^{6}} d x

x \to 0 + lim (x (ln (4 x))^{2})

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{x}

\frac{d}{d x} (9 x^{2} - 2 x + 1)