Author Topic: intergration q's (Read 3010 times) Tweet Share

darkmaster25 · « **on:** July 01, 2012, 05:41:50 pm »

I don't have solutions to this but what is everyone else getting for : $\int \frac{3x+4}{x^2-4x+13}$
I'm getting log_e(x^{2}-4x+13)+(8/3)tan^-1((x-2)/3) and according to my calculator i am wrong. why?

paulsterio · « **Reply #1 on:** July 01, 2012, 05:53:06 pm »

What answer does your calculator give?

b^3 · « **Reply #2 on:** July 01, 2012, 06:00:11 pm »

This is what I got as an answer $\frac{3}{2}\ln\left(x^{2}-4x+13\right)+\frac{10}{3}\tan^{-1}\left(\frac{x-2}{3}\right)+C$ .
Basically first trying to make the top a multiple of the derivative of the bottom, so that we can use substitution, then from that split the integral up, integrate the first part using the substitution, and integrate the second part by completing the square for the bottom, so that you can get it in the form $\frac{a}{x^2+a^2}$ to get a $\tan^{-1}(\frac{x}{a})$ out, then sub the u back in and add the +C.

I've done the working below, but maybe have a try first from whats above before having a look at it. Anyway hope it helps

$\begin{alignedat}{1}\int\frac{3x+4}{x^{2}-4x+13}dx & =\int\frac{3x-6+6+4}{x^{2}-4x+13}dx \\ & =\int\frac{3x-6}{x^{2}-4x+13}dx+\int\frac{10}{x^{2}-4x+13}dx \\ \mathrm{Let\:}u=x^{2}-4x+13 \\ \frac{du}{dx}=2x-4 \\ \int\frac{3x+4}{x^{2}-4x+13}dx & =\int\frac{3x-6}{u}\frac{1}{2x-4}du+\int\frac{10}{x^{2}-4x+4-4+13}dx \\ & =\int\frac{3}{2u}du+\int\frac{10}{(x-2)^{2}+9}dx \\ & =\int\frac{3}{2u}du+\frac{10}{3}\int\frac{3}{(x-2)^{2}+9}dx \\ & =\frac{3}{2}\ln|u|+\frac{10}{3}\tan^{-1}\left(\frac{x-2}{3}\right)+C \\ & =\frac{3}{2}\ln\left(x^{2}-4x+13\right)+\frac{10}{3}\tan^{-1}\left(\frac{x-2}{3}\right)+C\:\left(\mathrm{As\:}x^{2}-4x+13>0\:\mathrm{for\:all\:}x\right) \end{alignedat}$

EDIT: Added the "for all x" into the last line.

darkmaster25 · « **Reply #3 on:** July 01, 2012, 06:04:10 pm »

darkmaster25 · « **Reply #4 on:** July 01, 2012, 06:06:09 pm »

okay i see my mistake... thanks

darkmaster25 · « **Reply #5 on:** July 01, 2012, 08:56:00 pm »

Another question: do we have to put in the absolute value in logs. For example Ln would Ln(x^2-4x+13) require absolute value signs. When do we use or not use the absolute value sign. thanks.

Jenny_2108 · « **Reply #6 on:** July 01, 2012, 09:00:23 pm »

Normally I put absolute value for integration of log but in this case, x^2-4x+13 is always > 0 so we dont have to put absolute value

Phy124 · « **Reply #7 on:** July 01, 2012, 09:01:52 pm »

You use () brackets when the contents of the log will be $> 0$ for all real numbers. Whilst you use the absolute value || when the contents of the log will be $\leq 0$ .

I would usually put || first and then change it to () if applicable.

Jenny_2108 · « **Reply #8 on:** July 01, 2012, 09:05:00 pm »

Quote from: The AN Dunce on July 01, 2012, 09:01:52 pm

You use () brackets when the contents of the log will be $> 0$ for all real numbers. Whilst you use the absolute value || when the contents of the log will be $\leq 0$ .

I would usually put || first and then change it to () if applicable.

yep, thats true because in log(a), a need to be greater than 0

Somye · « **Reply #9 on:** July 01, 2012, 09:08:49 pm »

When you integrate something, your log(e) should generally give you an absolute value sign.

Remember that the inside of any log should be greater than 0, as this is where the function is actually defined

So the absolute value signs make sure our value inside the log is > 0 and hence actually has the function as defined
When we look at $x^2 - 4x + 13$ , for all values of x, it outputs a value of > 0 (all values above x axis), and hence the natural log function can exist without the need for an absolute value

If we think about it in terms of composite functions, Range of $x^2 - 4x + 13$ has to be equal to or a subset of the Domain of $\log_{e}x$ . It is (Range is [9,infinity) and Domain is (0,infinity)), and hence absolute value signs aren't required

darkmaster25 · « **Reply #10 on:** July 02, 2012, 06:06:40 pm »

Cheers guys, I need help with another question, how would I integrate this $\int \frac{1}{1+sin(x)}$ dx

b^3 · « **Reply #11 on:** July 02, 2012, 06:39:24 pm »

Quote from: darkmaster25 on July 02, 2012, 06:06:40 pm

Cheers guys, I need help with another question, how would I integrate this $\int \frac{1}{1+sin(x)}$ dx

This is how I would approach it, (multiplying by the conjugate)
$\begin{alignedat}{1}\int\frac{1}{1+\sin(x)}dx & =\int\frac{1-\sin(x)}{1-\sin^{2}(x)}dx \\ & =\int\frac{1-\sin(x)}{\cos^{2}(x)}dx \\ & =\int\frac{1}{\cos^{2}(x)}dx-\int\frac{\sin(x)}{\cos^{2}(x)}dx \\ & =\int\sec^{2}(x)dx-\int\frac{\sin(x)}{\cos^{2}(x)}dx \end{alignedat} $
Then from there you should be able to integrate it (using the right substitution for the second part) to get $\tan(x)-\sec(x)+C$ which is the same as $\frac{\sin(x)-1}{\cos(x)}+C$

Somye · « **Reply #12 on:** July 02, 2012, 06:39:33 pm »

Multiply by conjugate and split it into two separate integrals

darkmaster25 · « **Reply #13 on:** July 02, 2012, 06:52:55 pm »

nice... how do you recognise what to do straight away, if there a trick or is it just practise?

pi · « **Reply #14 on:** July 02, 2012, 06:55:05 pm »

Quote from: darkmaster25 on July 02, 2012, 06:52:55 pm

nice... how do you recognise what to do straight away, if there a trick or is it just practise?

It's just practice

Recognising things like that takes time

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Author Topic: intergration q's (Read 3010 times) Tweet Share

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intergration q's

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