What is the integral of e^{4x-3} ?

The integral of e^{4x-3} is 1/4 e^{4x-3}+C

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integral of e^{4x-3}

\int e^{4 x - 3} d x

\frac{1}{4} e^{4 x - 3} + C

Solution steps

\int e^{4 x - 3} 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 - 3

= \frac{1}{4} e^{4 x - 3}

Add a constant to the solution

= \frac{1}{4} e^{4 x - 3} + C

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

x \to 2 lim (∣ x ∣)

x \to 0 lim (x + 5)

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

n = 3 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}