What is the integral of e^{(4x)/3}(4/3) ?

The integral of e^{(4x)/3}(4/3) is e^{(4x)/3}+C

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integral of e^{(4x)/3}(4/3)

\int e^{\frac{4 x}{3}} (\frac{4}{3}) d x

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

Solution steps

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

Take the constant out:

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

= \frac{4}{3} \cdot \int e^{\frac{4 x}{3}} d x

Apply u-substitution

: \int e^{u} \frac{3}{4} d u

= \frac{4}{3} \cdot \int e^{u} \frac{3}{4} d u

Take the constant out:

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

= \frac{4}{3} \cdot \frac{3}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{4}{3} \cdot \frac{3}{4} e^{u}

Substitute back

u = \frac{4 x}{3}

= \frac{4}{3} \cdot \frac{3}{4} e^{\frac{4 x}{3}}

Simplify

\frac{4}{3} \cdot \frac{3}{4} e^{\frac{4 x}{3}} : e^{\frac{4 x}{3}}

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

Add a constant to the solution

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

- \frac{3 y}{3 x + 2 y}

\int 41 tan^{5} (x) sec^{4} (x) d x

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 2 y = 0

10 y^{^{''}} + 1000 y = 140 cos (8 t)

What is the integral of e^{(4x)/3}(4/3) ?
The integral of e^{(4x)/3}(4/3) is e^{(4x)/3}+C