Author Topic: Interesting questions (spesh) (Read 11829 times) Tweet Share

m@tty · « **Reply #15 on:** September 27, 2010, 10:11:47 pm »

Quote from: Martoman on September 27, 2010, 10:01:52 pm

yep thats right.... was eulers necessary?

That's what I was thinking... xD

Martoman · « **Reply #16 on:** September 27, 2010, 10:15:21 pm »

Quote from: brightsky on September 27, 2010, 10:10:06 pm

$2 (\sqrt{x + 4} + \log_e|\sqrt{x+4} - 2| - \log_e|\sqrt{x+4} - 2| + C$

A plus sign instead of negative between the the $\sqrt{x+4}-2$ on the second log term otherwise fabulous.

brightsky · « **Reply #17 on:** September 27, 2010, 10:20:42 pm »

Continuing Martoman's thread of awesome questions:

$\text{1.} \int \sqrt{2x - x^2} dx$

$\text{2.} \int \frac{x^2 + x + 1}{(x+1)^2 (x-2)} dx$

$\text{3.} \int^{1}_{0} \frac{\arctan(x)}{\log_e(x+1)} dx$

m@tty · « **Reply #18 on:** September 27, 2010, 10:21:23 pm »

brightsky, why aren't you doing spesh now? You'd own it nao.

EDIT:

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

Let $\frac{x^2+x+1}{(x+1)^2(x-2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-2}$

$x^2+x+1=A(x+1)(x-2)+B(x-2)+C(x+1)^2$

$x^2+x+1=x^2(A+C)+x(-A+B+2C)+(-2A-2B+C)$

Equating coefficients:

$A+C=1$ ...1 ; $-A+B+2C=1$ ...2 ; $-2A-2B+C=1$ ...3

1->2... $-(A+C)+B+3C=-1+B+3C=1 \implies B+3C=2$ ...4

1->3... $-2(A+C)-2B+3C=-2-2B+3C=1 \implies -2B+3C=3$ ...5

4->5... $-2(B+3C)+9C=-4+9C=3 \implies C=\frac{7}{9}$

Sub $C=\frac{7}{9}$ into 5

$-2B+3\cdot \frac{7}{9}=3 \implies B=-\frac{1}{3}$

Sub $C=\frac{7}{9}$ into 1

$A+\frac{7}{9}=1\implies A=\frac{2}{9}$

$\therefore \frac{x^2+x+1}{(x+1)^2(x-2)}=\frac{2}{9(x+1)}-\frac{1}{3(x+1)^2}+\frac{7}{9(x-2)}$

$\int \frac{x^2+x+1}{(x+1)^2(x-2)}dx=\frac{2}{9}\log_e|x+1|+\frac{1}{3(x+1)}+\frac{7}{9}\log_e|x-2|+C$

Martoman · « **Reply #19 on:** September 27, 2010, 10:35:42 pm »

Spoliers abound!~

1)

Let $u = \sqrt{2x-x^2}$

$u^2 = 2x-x^2$

$u^2 = -(x-1)^2+1$

$\sqrt{1-u^2} + 1 = x$

$dx = \frac{-u}{\sqrt{1-u^2}}du$

$\int \frac{-u^2}{\sqrt{1-u^2}}du$ .... then similar reasoning with my 2nd question.

2) Use awesome partial fractions

sajib_mostofa · « **Reply #20 on:** September 27, 2010, 10:56:39 pm »

2)

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

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

$\therefore x^2+x+1 = A(x+1)(x-2)+B(x-2)+C(x+1)^2$

$x\rightarrow2, C = \frac{7}{9}$

$x\rightarrow-1, B = \frac{-1}{3}$

$x\rightarrow3, A = \frac{2}{9}$ , (subbing the values for B and C found previously)

$\therefore\frac{x^2+x+1}{(x+1)^2(x-2))} =\frac{2}{9}\int \frac{1}{x+1}dx-\frac{1}{3}\int \frac{1}{(x+1)^2}dx + \frac{7}{9}\int \frac{1}{(x-2)} dx$

$= \frac{2}{9}\left|log_e(x+1)\right| + \frac{1}{3(x+1)}+ \frac{7}{9}\left|log_e(x-2)\right| +c<br />$

Martoman · « **Reply #21 on:** September 27, 2010, 11:03:37 pm »

Yep that corresponds to my thinking

brightsky · « **Reply #22 on:** September 27, 2010, 11:04:43 pm »

Quote from: Martoman on September 27, 2010, 10:35:42 pm

Spoliers abound!~

1)

Let $u = \sqrt{2x-x^2}$

$u^2 = 2x-x^2$

$u^2 = -(x-1)^2+1$

$\sqrt{1-u^2} + 1 = x$

$dx = \frac{-u}{\sqrt{1-u^2}}du$

$\int \frac{-u^2}{\sqrt{1-u^2}}du$ .... then similar reasoning with my 2nd question.

2) Use awesome partial fractions

Brilliant as always. There is a much more painful way but your way is much neater.

Quote from: sajib_mostofa on September 27, 2010, 10:56:39 pm

Since I cant be bothered typing it out, is the answer for 2):

$\frac{2}{9}log_e(x+1) +\frac{7}{9}log_e(x-2) +\frac{1}{3(x+1)}$

Yep, + C :p. Strictly speaking, the logs should bear absolute value signs.

Martoman · « **Reply #23 on:** September 27, 2010, 11:15:49 pm »

Quote from: brightsky on September 27, 2010, 11:04:43 pm

Quote from: Martoman on September 27, 2010, 10:35:42 pm
Spoliers abound!~

1)

Let $u = \sqrt{2x-x^2}$

$u^2 = 2x-x^2$

$u^2 = -(x-1)^2+1$

$\sqrt{1-u^2} + 1 = x$

$dx = \frac{-u}{\sqrt{1-u^2}}du$

$\int \frac{-u^2}{\sqrt{1-u^2}}du$ .... then similar reasoning with my 2nd question.

2) Use awesome partial fractions

Brilliant as always. There is a much more painful way but your way is much neater.

Quote from: sajib_mostofa on September 27, 2010, 10:56:39 pm
Since I cant be bothered typing it out, is the answer for 2):

$\frac{2}{9}log_e(x+1) +\frac{7}{9}log_e(x-2) +\frac{1}{3(x+1)}$

Yep, + C :p. Strictly speaking, the logs should bear absolute value signs.

what are the general steps to this painful method?

Martoman · « **Reply #24 on:** September 27, 2010, 11:22:21 pm »

I think brightsky meant my answer to the first question...

sajib_mostofa · « **Reply #25 on:** September 27, 2010, 11:24:19 pm »

Quote from: Martoman on September 27, 2010, 11:22:21 pm

I think brightsky meant my answer to the first question...

ahh ok haha, my mistake

brightsky · « **Reply #26 on:** September 27, 2010, 11:26:55 pm »

Quote

what are the general steps to this painful method?

Using the completing the square technique on $2x^2 - x$ , substituting u = whatever is in the (..)^2, then doing some crazy trigonometric substitution that no one would've thought of after that. Even then the integrand is majorly complicated.

Martoman · « **Reply #27 on:** September 27, 2010, 11:32:00 pm »

I like being a simpleton then :smitten:

and i've spent the last 20 mins trying that last one but to no avail in terms of integration. I bet it has something to do with the terminals... gah.. if i had a point that would be lovely :knuppel2:

sajib_mostofa · « **Reply #28 on:** September 28, 2010, 12:01:50 am »

I've provided some working out for 2) in my edited post as I had nothing better to do

Martoman · « **Reply #29 on:** September 28, 2010, 12:04:50 am »

lolll I'm facinated with q 3 still but have to go to bedd :buck2:

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