What is the integral of e^{2-5x} ?

The integral of e^{2-5x} is -1/5 e^{2-5x}+C

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integral of e^{2-5x}

\int e^{2 - 5 x} d x

- \frac{1}{5} e^{2 - 5 x} + C

Solution steps

\int e^{2 - 5 x} d x

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= \int - \frac{1}{5} e^{u} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{5} e^{u}

Substitute back

u = 2 - 5 x

= - \frac{1}{5} e^{2 - 5 x}

Add a constant to the solution

= - \frac{1}{5} e^{2 - 5 x} + C

x \to 2 + lim (\frac{1}{x ^{2} - 5 x + 6})

2 y^{^{''}} + (a + 1) y = 0

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

y = 1 + 3 x^{2} (ln (x^{2}))

f (x) = (x^{3} + 4) e^{x}