Raising/lowering with natural logs

69 Views Asked by At

I had a question on a test, and while I have already figured out that I should have done u substitution (I was running out of time and my brain froze), I was wondering if the following would be legal?

The question is $\int \frac{4 \ln x + 5}{x} dx$

(again, u sub... I know).

However, I was wondering if I could raise the top half of the fraction without doing the same to the bottom? Is this even acceptable in any way/shape/form?

eg $\int \frac{e^4 x + e^5}{x}dx$

or can I only do this for equations??

2

There are 2 best solutions below

1
On BEST ANSWER

I liked your question: Is this even acceptable in any way/shape/form?
It suggests that you already have a hunch that the answer is: No, this is not acceptable in any way/shape/form.

It is not acceptable to simply change the numerator of a fraction by taking e to the power of that numerator.

Even if you had an equation and took e to the power of both sides of the equation, it would not be acceptable to apply that only to the numerator of the fraction. You would need to take e to the power of the fraction as a whole.

Even if you had an equation, took e to the power of both sides of the equation, and took e to the power of the fraction as a whole, it would not be acceptable to "distribute" that action to the numerator and denominator separately. For example, $4^{\frac{1}{2}} ≠ \frac{4^1}{4^2}$.

Finally, even if you did have justification to take e to the power of $(4\ln x + 5)$, it would not simplify to $e^4 x + e^5$. Instead, you would do this:

$e^{4\ln x + 5} =$
$e^{4\ln x} * e^5 =$
$e^{\ln(x^4)} * e^5 =$
$x^4 * e^5 =$
$e^5 * x^4$

I hope that helps.
One good tip: if there is "$\ln x$" in your integral, then whether you do u-substitution or integration by parts, you almost certainly want "u" to equal something that includes the $\ln x$. In your case, $u = 4\ln x + 5$ is the best choice.

0
On

If you are integrating a function $\int f(x)\, dx$, then $\int e^{f(x)}\, dx$ is a completely different function. For example $$ \int \ln x\, dx=x\ln x-x+C. $$ But, $$ \int e^{\ln x}\,dx=\int x\, dx=\frac{1}{2}x^2+C. $$ Similarly $$ \int \frac{\ln x}{x}\, dx=\frac{1}{2}(\ln x)^2+C. $$ But, $$ \int \frac{e^{\ln x}}{x}\, dx=\int \frac{x}{x}\, dx=x+C $$ and $$ \int e^{\frac{\ln x}{x}}\,dx =\int x^{\frac{1}{x}}\,dx. $$