What is the integral of e^{7x+2} ?

The integral of e^{7x+2} is 1/7 e^{7x+2}+C

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integral of e^{7x+2}

\int e^{7 x + 2} d x

\frac{1}{7} e^{7 x + 2} + C

Solution steps

\int e^{7 x + 2} 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 + 2

= \frac{1}{7} e^{7 x + 2}

Add a constant to the solution

= \frac{1}{7} e^{7 x + 2} + C

\frac{x ^{6} + e ^{x}}{x + 3}

\frac{d}{d x} (\frac{1}{x + y})

\frac{d}{d x} ((3 x^{2} + 2)^{2} (x^{2} - 5 x)^{3})

(1 - x) (x^{2} - 7)^{2}

y = \frac{9}{x ^{2} + 5}, (2, 1)