integration help? (1 Viewer)

[nfreidman]

how do you integrate

10/[(x-1)(x²+9)] ?

thanks in advance

 

[deswa1]

Have you tried splitting into partial fractions?

 

[barbernator]

the answer is ln(x-1) -1/2ln(x^2+9) - 1/3tan-1(x/3) + C post working in a sec

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{10}{(x-1)(x^{2}+9)}\\ = \frac{a}{x-1}@plus;\frac{bx@plus;c}{x^2@plus;9}\\ =\frac{a(x^2@plus;9)@plus;(bx@plus;c)(x-1)}{(x-1)(x^2-9)}\\ taking, numerator, and, expanding, and, factorising\\ =(a@plus;b)x^2@plus;(c-b)x@plus;(9a-c)= 10\\ we now have 3 equations\\ a@plus;b=0 (1)\\ c-b@plus;0 (2)\\ 9a-c=10(3)\\ solving them simultaneously we get a=1,b=-1,c=-1\\ we now have our partial fractions\\ \frac{1}{x-1}-\frac{x}{x^2@plus;9}-\frac{1}{x^2@plus;9}\\ integrating\\ \int \frac{1}{x-1}-\int \frac{x}{x^2@plus;9}-\int \frac{1}{x^2@plus;9} \\ = ln(x-1)-\frac{1}{2}ln(x^2@plus;9)-\frac{1}{3}tan^-1(\frac{x}{3})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{10}{(x-1)(x^{2}-9)}\\ = \frac{a}{x-1}+\frac{bx+c}{x^2+9}\\ =\frac{a(x^2+9)+(bx+c)(x-1)}{(x-1)(x^2-9)}\\ taking numerator and expanding and factorising\\ =(a+b)x^2+(c-b)x+(9a-c)= 10\\ we now have 3 equations\\ a+b=0 (1)\\ c-b+0 (2)\\ 9a-c=10(3)\\ solving them simultaneously we get a=1,b=-1,c=-1\\ we now have our partial fractions\\ \frac{1}{x-1}-\frac{x}{x^2+9}-\frac{1}{x^2+9}\\ integrating\\ \int \frac{1}{x-1}-\int \frac{x}{x^2+9}-\int \frac{1}{x^2+9} \\ = ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^-1(\frac{x}{3})" title="\frac{10}{(x-1)(x^{2}-9)}\\ = \frac{a}{x-1}+\frac{bx+c}{x^2+9}\\ =\frac{a(x^2+9)+(bx+c)(x-1)}{(x-1)(x^2-9)}\\ taking numerator and expanding and factorising\\ =(a+b)x^2+(c-b)x+(9a-c)= 10\\ we now have 3 equations\\ a+b=0 (1)\\ c-b+0 (2)\\ 9a-c=10(3)\\ solving them simultaneously we get a=1,b=-1,c=-1\\ we now have our partial fractions\\ \frac{1}{x-1}-\frac{x}{x^2+9}-\frac{1}{x^2+9}\\ integrating\\ \int \frac{1}{x-1}-\int \frac{x}{x^2+9}-\int \frac{1}{x^2+9} \\ = ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^-1(\frac{x}{3})"+C /></a>

+C

if that is wrong, someone please tell. Also, i have no idea how to do spaces in latex

 

[SpiralFlex]

Where's your infinitesimally small increment of the variable x boy?

 

[barbernator]

Where's your infinitesimally small increment of the variable x boy?

mi scusi

 

[nfreidman]

if that is wrong, someone please tell. Also, i have no idea how to do spaces in latex[/QUOTE]

i think it's all correct except for one minus sign you missed in the question, but i can still use your working to solve my question so thank you

could you explain how you knew that the terms on the top were A with that particular denominator, and Bx+c with that denominator?

 

[Siddy123]

could you explain how you knew that the terms on the top were A with that particular denominator, and Bx+c with that denominator?[/QUOTE]


principle of splitting using partial fractions...

 

[SpiralFlex]

Alternatively, you could use the Oliver's Heaviside Cover Up method. That is if you want to speed up your working.

 

[barbernator]

Alternatively, you could use the Oliver's Heaviside Cover Up method. That is if you want to speed up your working.

doesn't that only work with linear factors in the denominator?

 

[SpiralFlex]

doesn't that only work with linear factors in the denominator?

If you had a linear and an unfactorisable, you could use it for one, then you need to use the traditional methods for the other. Works better for linear.

 

[barbernator]

could you explain how you knew that the terms on the top were A with that particular denominator, and Bx+c with that denominator?

we know from this fraction, that it cannot be simplified without splitting it up, i.e the degree of the numerator is less than the denominator. Therefore, this fraction must come from an addition of two fractions. There is a possibility that the second fraction may end up only having a constant term on top as well once we have resolved the fractions, yet we know that the first fraction cannot have a linear factor as the numerator, as we would have been able to simplify the original fraction by division first.

Thats not really the best explanation, spiralflex will confirm or deny

 

[qwerty44]

we know from this fraction, that it cannot be simplified without splitting it up, i.e the degree of the numerator is less than the denominator. Therefore, this fraction must come from an addition of two fractions. There is a possibility that the second fraction may end up only having a constant term on top as well once we have resolved the fractions, yet we know that the first fraction cannot have a linear factor as the numerator, as we would have been able to simplify the original fraction by division first.

Thats not really the best explanation, spiralflex will confirm or deny

I cant really help with the Ext 2 stuff. But a space in latex is made by ~ between words.

 

[nightweaver066]

I cant really help with the Ext 2 stuff. But a space in latex is made by ~ between words.

:O Really. I've always used $ $ to type text.

 

[kfnmpah]

no one should ever ask maths questions on here, especially not integration, when you can just go into wolframalpha.com and type it in and click 'show steps'

duh.

 

[barbernator]

no one should ever ask maths questions on here, especially not integration, when you can just go into wolframalpha.com and type it in and click 'show steps'

duh.

what do you live for?

 

[Nooblet94]

no one should ever ask maths questions on here, especially not integration, when you can just go into wolframalpha.com and type it in and click 'show steps'

duh.

Lots of the time wolframalpha uses some ridiculously long method that utilises stuff that's way beyond the HSC course. But I agree in that people should check their answer their first and then post here.