What is the integral of e^{4x+1} ?

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

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integral of e^{4x+1}

\int e^{4 x + 1} d x

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

Solution steps

\int e^{4 x + 1} 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 + 1

= \frac{1}{4} e^{4 x + 1}

Add a constant to the solution

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

y^{^{''}} + 20 y^{^{'}} + 500 y = 12

\int \frac{1}{cos ^{2} ( x )} d x

x \to 2 lim (\frac{x ^{3} + 7 x + 6}{x ^{3} + x ^{2} - 4 x - 4})

\int \frac{1}{2 ^{2} + 3 x ^{2}} d x

f (x) = 2 x^{3} - 24 x^{2} + 97 x + 7