What is the integral of $e^{2 x}$ $e^(2x)$ ?

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Answer 1 · 2018-05-02T21:06:50

The answer

$\int e^{2 x} \cdot d x = \frac{1}{2} [e^{2 x}] + c$ $inte^(2x)*dx=1/2[e^(2x)]+c$

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$\int e^{2 x} \cdot d x = \frac{1}{2} \int 2 \cdot e^{2 x} \cdot d x = \frac{1}{2} [e^{2 x}] + c$ $inte^(2x)*dx=1/2int2*e^(2x)*dx=1/2[e^(2x)]+c$

Answer 2 · 2018-05-03T00:29:57

$\frac{1}{2} e^{2 x} + C$ $1/2e^(2x)+C$

Given: $inte^(2x) \ dx$ .

We can manipulate as follows:

$inte^(2x) \ dx$

$=int1/2*2e^(2x) \ dx$

$=1/2int2e^(2x) \ dx$

Now, let $u=2x,:.du=2 \ dx,dx=(du)/2$

$=1/2int2e^u*(du)/2$

$=1/2inte^u \ du$

$=1/2e^u+C$

Replace back $u=2x$ to get the final integral:

$=1/2e^(2x)+C$

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