Integration Tings

[hi.itsjessi]

Hi friends! Just a thing that's gotten me a little perplexed.

When we're required to integrate a fraction, how do we know whether to simplify the question by dragging some stuff outside of the integral and then bringing the denominator up, or whether to look at the question in terms of logs?

[RuiAce]

I'm not sure what's going on with the first part here. What constitutes a question that involves "dragging some stuff outside of the integral and then bringing the denominator up"?

[hi.itsjessi]

I'm not sure what's going on with the first part here. What constitutes a question that involves "dragging some stuff outside of the integral and then bringing the denominator up"?

Sorry for the confusion! What I meant by the first part was that if you're given a question like ∫ (-3/2x^2)dx, you can simplify it first by dragging -3/2 outside of the integral sign which gives you -3/2 ∫ 1/x^2 , and then you bring the denominator of whatever's inside the integral, up to make it an exponential (and then you can do whatever you need to, to solve the question  ). Whereas with something like ∫(5/3x)dx, you don't bring the 5/3 outside of the integral, but instead, you look at it in terms of logs.
My issue is that I can never tell which way I should approach the question, I'm actually terrible with maths 

[jamonwindeyer]

Sorry for the confusion! What I meant by the first part was that if you're given a question like ∫ (-3/2x^2)dx, you can simplify it first by dragging -3/2 outside of the integral sign which gives you -3/2 ∫ 1/x^2 , and then you bring the denominator of whatever's inside the integral, up to make it an exponential (and then you can do whatever you need to, to solve the question  ). Whereas with something like ∫(5/3x)dx, you don't bring the 5/3 outside of the integral, but instead, you look at it in terms of logs.
My issue is that I can never tell which way I should approach the question, I'm actually terrible with maths 

Hey! Log integrals are always of the form derivative on top, function on bottom. So, look for a power of \(x\) on top that is one less than the bottom!

Alternatively, try doing it without logs first. If you can't, or it breaks, then you go to logs

[Opengangs]

Hey.

So with integration, you'd want to make things easier for you first! That typically means you should always bring any constants to the front where possible! So in your example:

\[ \int \frac{5}{3x}\,dx\]

You'd want to bring the constant 5/3 outside to give you:

\[ \frac{5}{3}\int \frac{1}{x}\,dx = \frac{5}{3}\ln x + C\]

Log integrals are often of the form: \(\int \frac{f'(x)}{f(x)}\,dx\), so like jamon stated, look for whether the top is one less power than the bottom. That's a hint that allows you to use the log property of integrals!

[RuiAce]

Sorry for the confusion! What I meant by the first part was that if you're given a question like ∫ (-3/2x^2)dx, you can simplify it first by dragging -3/2 outside of the integral sign which gives you -3/2 ∫ 1/x^2 , and then you bring the denominator of whatever's inside the integral, up to make it an exponential (and then you can do whatever you need to, to solve the question  ). Whereas with something like ∫(5/3x)dx, you don't bring the 5/3 outside of the integral, but instead, you look at it in terms of logs.
My issue is that I can never tell which way I should approach the question, I'm actually terrible with maths 

The question's mostly addressed above but I'm gonna make a remark. The main reason why that method works for the question you wrote down is essentially because it was of the form \( \int x^n\,dx \) where \(n \neq -1\). In your case, you had \( n = -2\).

So by quoting that rule, \( \int \frac{1}{x^2}\,dx = \int x^{-2}\,dx = -x^{-1} + C = -\frac1x + C\).

Whereas when \( n = -1\), we have \(\int x^{-1}\,dx = \int \frac1x \, dx = \ln(x) + C\). In general, when the derivative of what's on the bottom appears in the top, you have those integrals that go to a log. But for the other one there was no problem isolating a power on \(x\) all by itself, so you should be using the power law instead.

Note also (from the reference sheet) that the same thing can be done with integrals of the form \( \int (ax+b)^n\,dx\), for example, \( \int \frac{1}{(9x-1)^2}\,dx\).