What is the integral of 4e^{2x-1} ?

The integral of 4e^{2x-1} is 2e^{2x-1}+C

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integral of 4e^{2x-1}

\int 4 e^{2 x - 1} d x

2 e^{2 x - 1} + C

Solution steps

\int 4 e^{2 x - 1} d x

Take the constant out:

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

= 4 \cdot \int e^{2 x - 1} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= 4 \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x - 1

= 4 \cdot \frac{1}{2} e^{2 x - 1}

Simplify

4 \cdot \frac{1}{2} e^{2 x - 1} : 2 e^{2 x - 1}

= 2 e^{2 x - 1}

Add a constant to the solution

= 2 e^{2 x - 1} + C

\int_{- 1}^{1} \frac{1}{t + 1} d t

\frac{d y}{d x} = 42 y x^{6}

\int_{0}^{1} (2 x + y)^{8} d x

f (x) = x^{3} - \frac{3}{2} x^{2} + 4 x + 1

(x - 4) y^{^{'}} + y = x^{2} + 8