How do you integrate $e^{4 x} d x$ $e^(4x) dx$ ?

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How do you integrate $e^{4 x} d x$ $e^(4x) dx$ ?

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Answer 1 · 2016-06-24T16:58:23

$\frac{1}{4} e^{4 x} + C$ $1/4e^(4x)+C$

Explanation:

We will use the integration rule for $e^{x}$ $e^x$ :

$\int e^{u} d u = e^{u} + C$ $inte^udu=e^u+C$

So, for the given integral, let $u = 4 x$ $u=4x$ . This implies that $d u = 4 d x$ $du=4dx$ .

$\int e^{4 x} d x = \frac{1}{4} \int e^{4 x} \cdot 4 d x = \frac{1}{4} \int e^{u} d u = \frac{1}{4} e^{u} + C$ $inte^(4x)dx=1/4inte^(4x)*4dx=1/4inte^udu=1/4e^u+C$

Since $u = 4 x$ $u=4x$ :

$\frac{1}{4} e^{u} + C = \frac{1}{4} e^{4 x} + C$ $1/4e^u+C=1/4e^(4x)+C$

We can differentiate this answer to check that we get $e^{4 x}$ $e^(4x)$ . Indeed, through the chain rule, the $\frac{1}{4}$ $1/4$ we had to add gets "undone" by the $4$ $4$ coming from the power of $4 x$ $4x$ via the chain rule.

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How do you integrate $e^{4 x} d x$ $e^(4x) dx$ ?

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