# Integrate (2-3x)^-2

• Jan 18th 2013, 03:32 PM
Furyan
Integrate (2-3x)^-2
Hello

Would someone please tell me what method I should use to integrate the following.

$\displaystyle \int(2 - 3x)^{-2}$

I have tried writing the numerator as the derivative of the denominator, but then I'd have an f(x) to compensate for. I thought of using partial fractions, but haven't been able to write this as the sum of two fractions. I'm wondering if I have to use another method, substitution or by parts.

Thank you.
• Jan 18th 2013, 03:37 PM
Plato
Re: Integrate (2-3x)^-2
Quote:

Originally Posted by Furyan
Hello

Would someone please tell me what method I should use to integrate the following.

$\displaystyle \int(2 - 3x)^{-2}$

I have tried writing the numerator as the derivative of the denominator, but then I'd have an f(x) to compensate for. I thought of using partial fractions, but haven't been able to write this as the sum of two fractions. I'm wondering if I have to use another method, substitution or by parts.

What is the derivative of $\displaystyle \frac{(2-3x)^{-1}}{3}~?$

• Jan 18th 2013, 03:46 PM
Prove It
Re: Integrate (2-3x)^-2
Quote:

Originally Posted by Furyan
Hello

Would someone please tell me what method I should use to integrate the following.

$\displaystyle \int(2 - 3x)^{-2}$

I have tried writing the numerator as the derivative of the denominator, but then I'd have an f(x) to compensate for. I thought of using partial fractions, but haven't been able to write this as the sum of two fractions. I'm wondering if I have to use another method, substitution or by parts.

Thank you.

\displaystyle \displaystyle \begin{align*} \int{(2 - 3x)^{-2}\,dx} = -\frac{1}{3}\int{-3 (2 - 3x)^{-2}\,dx} \end{align*}

Now let \displaystyle \displaystyle \begin{align*} u = 2 - 3x \implies du = -3\,dx \end{align*} and the integral becomes \displaystyle \displaystyle \begin{align*} -\frac{1}{3}\int{u^{-2}\,du} \end{align*}. Go from here.
• Jan 18th 2013, 04:24 PM
Furyan
Re: Integrate (2-3x)^-2
Quote:

Originally Posted by Plato
What is the derivative of $\displaystyle \frac{(2-3x)^{-1}}{3}~?$

The derivative is $\displaystyle (2 - 3x)^{-2}$

So the answer is $\displaystyle \dfrac{1}{3}(2-3x)^{-1} + c$

Thank you so much. I was getting really confused and was thinking in terms of $\displaystyle \ln$ when I should have just been using the chain rule in reverse.
• Jan 18th 2013, 06:39 PM
Plato
Re: Integrate (2-3x)^-2
Quote:

Originally Posted by Furyan
The derivative is $\displaystyle (2 - 3x)^{-2}$
So the answer is $\displaystyle \dfrac{1}{3}(2-3x)^{-1} + c$
Thank you so mtergablesuch. I was getting really confused and was thinking in terms of $\displaystyle \ln$ when I should have just been using the chain rule in reverse.

Yes that my whole point!
I would tell students, forget rules, just know derivatives.
In indefinite integrable problems of what is it the derivative?

That is the whole and complete point!!

If you can find a function such its derivative is the given function, then you are done! Adding a $\displaystyle C$ actually means nothing to those of us who's research is on the integral.
• Jan 19th 2013, 01:07 AM
Prove It
Re: Integrate (2-3x)^-2
Quote:

Originally Posted by Plato
Yes that my whole point!
I would tell students, forget rules, just know derivatives.
In indefinite integrable problems of what is it the derivative?

That is the whole and complete point!!

If you can find a function such its derivative is the given function, then you are done! Adding a $\displaystyle C$ actually means nothing to those of us who's research is on the integral.

I'm sorry but I completely disagree with you. Yes, know that integration is essentially the opposite process to differentiation, but without knowing some rules like substitution, you are stuck trying to think of derivatives out of thin air. We have these rules, which are easily derived, to make our lives easier, not more difficult.
• Jan 19th 2013, 05:05 AM
Plato
Re: Integrate (2-3x)^-2
Quote:

Originally Posted by Prove It
I'm sorry but I completely disagree with you. Yes, know that integration is essentially the opposite process to differentiation, but without knowing some rules like substitution, you are stuck trying to think of derivatives out of thin air. We have these rules, which are easily derived, to make our lives easier, not more difficult.

If the point is "to make our lives easier, not more difficult" we should learn to use something like WolfranAlpha. There is even a cell phone app. In fact, I predict that within twenty years calculus textbooks will not have a section on Techniques of Integration. Given the tools available that have been for twenty five years (CAS, calculators, etc), it is hard to see why that is not already true.

Look at to OP. It is sad if one looks at that and does not see it as a derivative form. I find it truly sad that I have seen on this very board some who have to use a substitution to see $\displaystyle \int {\frac{{2xdx}}{{1 + x^4 }}}$ as an $\displaystyle \arctan$, even without any more effort.