Author Topic: Integration of factorised functions by hand? (Read 2081 times) Tweet Share

brightsky · « **Reply #15 on:** September 26, 2010, 06:35:55 pm »

Quote

I think you can... I've seen stuff like that on trials?
Although yeah ofcourse you cannot trust them.

Yep, you can, but it's a hell lot of work and (as TT might put it) involves a bit of wishful thinking. As for m@tty's one...even harder. :p

For those interested:

$\int \frac{1}{(3x^2 - 2x)^2} dx$

$= \int \frac{1}{x^2(3x-2)^2} dx$

Decompose the integrand into partial fractions. The partial fraction expansion is in the form: $\frac{A}{x^2} + \frac{B}{x} + \frac{C}{(3x-2)^2} + \frac{D}{(3x - 2)}$

So let $\frac{1}{x^2(3x-2)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x-2} + \frac{D}{(3x - 2)^2}$

$1 = Ax(3x - 2)^2 + B(3x - 2)^2 + Cx^2(3x - 2) + Dx^2$

$1 = Ax(9x^2 - 12x + 4) + B(9x^2 - 12x + 4) + Cx^2(3x - 2) + Dx^2$

$1 = 9Ax^3 - 12Ax^2 + 4Ax + 9Bx^2 - 12Bx + 4B + 3Cx^3 - 2Cx^2 + Dx^2$

$1 = x^3 (9A + 3C) + x^2 (9B - 12A - 2C + D) + x(4A - 12B) + (4B)$

Given that both sides are equal, that means,

$4B = 1$

$9A + 3C = 0$

$9B - 12A - 2C + D = 0$

$4A - 12B = 0$

Solving the set of simultaneous equations, we get $A = \frac{3}{4}$ , $B = \frac{1}{4}$ , $C = -\frac{9}{4}$ , $D = \frac{9}{4}$

Hence $\int \frac{1}{x^2(3x-2)^2} dx = \int \left(\frac{\frac{3}{4}}{x} + \frac{\frac{1}{4}}{x^2} + \frac{-\frac{9}{4}}{3x-2} + \frac{\frac{9}{4}}{(3x - 2)^2}\right) dx$

$= \int \left(\frac{3}{4x} + \frac{1}{4x^2} - \frac{9}{4(3x-2)} + \frac{9}{4(3x - 2)^2}\right) dx$

Integrating term by term:

$= \frac{3}{4} \log_e|x| - \frac{1}{4x} - \frac{3}{4} \log_e|3x - 2| - \frac{3}{4(3x - 4)} + C$

Hope this is right..

EDIT: Forgot the + C.

EDIT 2: Thanks Mao! Grr..keep doing that..:p

qshyrn · « **Reply #16 on:** September 26, 2010, 07:16:54 pm »

Quote from: brightsky on September 26, 2010, 06:35:55 pm

Quote

I think you can... I've seen stuff like that on trials?
Although yeah ofcourse you cannot trust them.

Yep, you can, but it's a hell lot of work and (as TT might put it) involves a bit of wishful thinking. As for m@tty's one...even harder. :p

For those interested:

$\int \frac{1}{(3x^2 - 2x)^2} dx$

$= \int \frac{1}{x^2(3x-2)^2} dx$

Decompose the integrand into partial fractions. The partial fraction expansion is in the form: $\frac{A}{x^2} + \frac{B}{x} + \frac{C}{(3x-2)^2} + \frac{D}{(3x - 2)}$

So let $\frac{1}{x^2(3x-2)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x-2} + \frac{D}{(3x - 2)^2}$

$1 = Ax(3x - 2)^2 + B(3x - 2)^2 + Cx^2(3x - 2) + Dx^2$

$1 = Ax(9x^2 - 12x + 4) + B(9x^2 - 12x + 4) + Cx^2(3x - 2) + Dx^2$

$1 = 9Ax^3 - 12Ax^2 + 4Ax + 9Bx^2 - 12Bx + 4B + 3Cx^3 - 2Cx^2 + Dx^2$

$1 = x^3 (9A + 3C) + x^2 (9B - 12A - 2C + D) + x(4A - 12B) + (4B)$

Given that both sides are equal, that means,

$4B = 1$

$9A + 3C = 0$

$9B - 12A - 2C + D = 0$

$4A - 12B = 0$

Solving the set of simultaneous equations, we get $A = \frac{3}{4}$ , $B = \frac{1}{4}$ , $C = -\frac{9}{4}$ , $D = \frac{9}{4}$

Hence $\int \frac{1}{x^2(3x-2)^2} dx = \int \left(\frac{\frac{3}{4}}{x} + \frac{\frac{1}{4}}{x^2} + \frac{-\frac{9}{4}}{3x-2} + \frac{\frac{9}{4}}{(3x - 2)^2}\right) dx$

$= \int \left(\frac{3}{4x} + \frac{1}{4x^2} - \frac{9}{4(3x-2)} + \frac{9}{4(3x - 2)^2}\right) dx$

Integrating term by term:

$= \frac{3}{4} \log_e|x| - \frac{1}{12x^3} - \frac{3}{4} \log_e|3x - 2| + \frac{(3x - 2)^3}{4}$

Hope this is right..

sorry first step aint right (3x^2-2x)^2 when expanded is not (9x^4-6x^3+4x^2)

brightsky · « **Reply #17 on:** September 26, 2010, 08:08:25 pm »

Quote from: 8039 on September 26, 2010, 04:43:46 pm

Quote from: 99.95_for_sure on September 26, 2010, 04:17:56 pm
$\int (2x-1)^{3}dx$

$=\frac{(2x-1)^{4}}{4*2}=\frac{(2x-1)^{4}}{8}$

$\int 4(2x+3)^{-2}dx$

$=\frac{4(2x+3)^{-1}}{-1*2}$

$=\frac{-2}{2x+3}$

What if it were something like (3x^2 - 2x)^-2?

Doing this one. :p

As for m@tty's one:

$\int \frac{1}{9x^4 - 6x^3 + 4x^2} dx$

$= \int \frac{1}{x^2(9x^2 - 6x + 4)} dx$

Hence the partial decomposition is in the form $\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{9x^2 - 6x + 4}$ since $9x^2 - 6x + 4$ is irreducible.

Thus let the integrand $\frac{1}{x^2(9x^2 - 6x + 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{9x^2 - 6x + 4}$

$1 = Ax(9x^2 - 6x + 4) + B(9x^2 - 6x + 4) + (Cx + D)x^2$

$1 = 9Ax^3 - 6Ax^2 + 4Ax + 9Bx^2 - 6Bx + 4B + Cx^3 + Dx^2$

$1 = x^3(9A + C) + x^2(9B - 6A + D) + x(4A - 6B) + (4B)$

Equating both sides, we get:

$4B = 1$

$9A + C = 0$

$9B - 6A + D = 0$

$4A - 6B = 0$

Solving the set of simultaneous equations, we get $A = \frac{3}{8}$ , $B = \frac{1}{4}$ , $C = -\frac{27}{8}$ , $D = 0$

Hence $\int \frac{1}{x^2(9x^2 - 6x + 4)} dx = \int \left(\frac{\frac{3}{8}}{x} + \frac{\frac{1}{4}}{x^2} + \frac{-\frac{27}{8}x}{9x^2 - 6x + 4}\right) dx$

$= \int \left(\frac{3}{8x} + \frac{1}{4x^2} - \frac{27x}{8(9x^2 - 6x + 4)}\right) dx$

Integrating term by term:

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \int \frac{27x}{8(9x^2 - 6x + 4)} dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{27}{8} \int \frac{x}{9x^2 - 6x + 4} dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{27}{8} \int \left(\frac{1}{3(9x^2 - 6x + 4)} + \frac{18x - 6}{18(9x^2 - 6x + 4)}\right) dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{27}{8} \cdot \frac{1}{3} \int \frac{1}{9x^2 - 6x + 4} dx - \frac{27}{8} \cdot \frac{1}{18} \int \frac{18x - 6}{9x^2 - 6x + 4} dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{9}{8} \int \frac{1}{9x^2 - 6x + 4} dx - \frac{3}{16} \int \frac{18x - 6}{9x^2 - 6x + 4} dx$

Substitute $u = 9x^2 - 6x + 4$ , $du = \frac{18x - 6} dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{9}{8} \int \frac{1}{9x^2 - 6x + 4} dx - \frac{3}{16} \int \frac{1}{u} du$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{9}{8} \int \frac{1}{9x^2 - 6x + 4} dx - \frac{3}{16} \log_e|u|$

Substitute back $u = 9x^2 - 6x + 4$ :

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{9}{8} \int \frac{1}{9x^2 - 6x + 4} dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{9}{8} \int \frac{1}{(3x - 1)^2 + 3} dx$

Substitute $v = 3x - 1$ , $dv = 3 dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{9}{8} \cdot \frac{1}{3} \int \frac{1}{v^2 + 3} dx$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{3}{8} \int \frac{1}{v^2 + 3} dx$

We know that $\int \frac{a}{x^2 + a^2} dx = \arctan\left(\frac{x}{a}\right) + C$

Hence $\int \frac{1}{v^2 + 3} dx = \int \frac{1}{v^2 + (\sqrt{3})^2} dx = \frac{\arctan\left(\frac{v^2}{\sqrt{3}}\right)}{\sqrt{3}} + C$

So the original equation becomes:

$\frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{3}{8} \frac{\arctan\left(\frac{v^2}{\sqrt{3}}\right)}{\sqrt{3}} + C$

Substitute $v = 3x - 1$ back into the equation:

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{3}{8} \frac{\arctan\left(\frac{(3x - 1)^2}{\sqrt{3}}\right)}{\sqrt{3}} + C$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{3\sqrt{3}}{8 \cdot 3} \arctan\left(\frac{(3x - 1)^2}{\sqrt{3}}\right) + C$

$= \frac{3}{8} \log_e|x| - \frac{1}{4x} - \frac{3}{16} \log_e|9x^2 - 6x + 4| - \frac{\sqrt{3}}{8} \arctan\left(\frac{(3x -1)^2}{\sqrt{3}}\right) + C$

EDIT: Thanks Mao.

Mao · « **Reply #18 on:** September 26, 2010, 08:20:23 pm »

Quote from: brightsky on September 26, 2010, 06:35:55 pm

$= \int \left(\frac{3}{4x} + \frac{1}{4x^2} - \frac{9}{4(3x-2)} + \frac{9}{4(3x - 2)^2}\right) dx$

Integrating term by term:

$= \frac{3}{4} \log_e|x| - \frac{1}{12x^3} - \frac{3}{4} \log_e|3x - 2| + \frac{(3x - 2)^3}{4}$

Hope this is right..

Buddy... you were going so well, what happened in the last step!

[everything except the last line is correct]

8039 · « **Reply #19 on:** September 26, 2010, 11:54:59 pm »

Quote from: brightsky on September 26, 2010, 06:35:55 pm

Quote

I think you can... I've seen stuff like that on trials?
Although yeah ofcourse you cannot trust them.

Yep, you can, but it's a hell lot of work and (as TT might put it) involves a bit of wishful thinking. As for m@tty's one...even harder. :p

For those interested:

$\int \frac{1}{(3x^2 - 2x)^2} dx$

$= \int \frac{1}{x^2(3x-2)^2} dx$

Decompose the integrand into partial fractions. The partial fraction expansion is in the form: $\frac{A}{x^2} + \frac{B}{x} + \frac{C}{(3x-2)^2} + \frac{D}{(3x - 2)}$

So let $\frac{1}{x^2(3x-2)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x-2} + \frac{D}{(3x - 2)^2}$

$1 = Ax(3x - 2)^2 + B(3x - 2)^2 + Cx^2(3x - 2) + Dx^2$

$1 = Ax(9x^2 - 12x + 4) + B(9x^2 - 12x + 4) + Cx^2(3x - 2) + Dx^2$

$1 = 9Ax^3 - 12Ax^2 + 4Ax + 9Bx^2 - 12Bx + 4B + 3Cx^3 - 2Cx^2 + Dx^2$

$1 = x^3 (9A + 3C) + x^2 (9B - 12A - 2C + D) + x(4A - 12B) + (4B)$

Given that both sides are equal, that means,

$4B = 1$

$9A + 3C = 0$

$9B - 12A - 2C + D = 0$

$4A - 12B = 0$

Solving the set of simultaneous equations, we get $A = \frac{3}{4}$ , $B = \frac{1}{4}$ , $C = -\frac{9}{4}$ , $D = \frac{9}{4}$

Hence $\int \frac{1}{x^2(3x-2)^2} dx = \int \left(\frac{\frac{3}{4}}{x} + \frac{\frac{1}{4}}{x^2} + \frac{-\frac{9}{4}}{3x-2} + \frac{\frac{9}{4}}{(3x - 2)^2}\right) dx$

$= \int \left(\frac{3}{4x} + \frac{1}{4x^2} - \frac{9}{4(3x-2)} + \frac{9}{4(3x - 2)^2}\right) dx$

Integrating term by term:

$= \frac{3}{4} \log_e|x| - \frac{1}{12x^3} - \frac{3}{4} \log_e|3x - 2| + \frac{(3x - 2)^3}{4} + C$

Hope this is right..

EDIT: Forgot the + C.

Awesome thanks!

Imagine doing all that work on an exam only to forget the +C

m@tty · « **Reply #20 on:** September 27, 2010, 12:21:48 am »

Haha, good work brightsky.

Quote from: 8039 on September 26, 2010, 11:54:59 pm

Awesome thanks!

Imagine doing all that work on an exam only to forget the +C

Know that you will NOT get anything like that in a methods exam. Or Spesh for that matter, that is way too much working for a VCE exam...

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Author Topic: Integration of factorised functions by hand? (Read 2081 times) Tweet Share

brightsky

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qshyrn

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brightsky

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Mao

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8039

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m@tty

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