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For example:

$$\int a^x \, dx = \int e^{x \ln a } \, dx$$ Let $u= (\ln a) x$

From here: $$ \frac{du}{dx} = \ln a $$ $$ \frac{1}{\ln a} du= dx $$

$$\int e^{x \ln a } \, dx = \int e^u \, du = \frac{1}{\ln a} e^u + C = \frac{1}{\ln a} a^x + C $$

This uses du/dx as a fraction, how can you change the base of an integral without doing this?

In the question Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? many people say $dy/dx$ is not a fraction.

user716881
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    Have you read all the answers of the linked question? If yes, is something unclear at these answers? – callculus42 Jun 03 '21 at 14:55
  • You're right to question why we are allowed to treat $dy/dx$ as a fraction in this case. Here, it is acceptable because we are just using the reverse chain rule in disguise, as explained in the question below. – Joe Jun 03 '21 at 14:57

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Well, you just have to use the definition of the substitution rule that says

$$\int_a^b f(\varphi(x))\varphi'(x) dx = \int_{\varphi(a)}^{\varphi(b)}f(u)du$$

In your example, $u=\varphi(x)=\ln(a)x$, so you know

$$ \int e^{x\ln{a}}\cdot \ln(a) dx = \int f(u) du$$

what means nothing else than $$ \int e^{x\ln{a}}dx = \int f(u) \frac{du}{\ln a}$$

HyperPro
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