Integration question from terry lee :( (1 Viewer)

coeyz · Apr 9, 2009

1. integrate (1-2x) / x^2 + 2x + 3 dx
2. integrate square root [ (x-1) / (x+1) ] dx

thanksssssss

shinn · Apr 10, 2009

$1)\ \int \frac{1-2x}{x^2 + 2x + 3}dx \\= \int \frac{3}{x^2 + 2x + 3}dx -\int \frac{2x+2}{x^2 + 2x + 3}dx \\= 3\int \frac{1}{(x+1)^2+2}dx - ln (x^2 + 2x +3) \\= \frac{3 \sqrt{2}}{2}\tan^{-1} (\frac{x+1}{\sqrt{2}}) - ln (x^2 + 2x +3) + C$

edit: yea my bad, didn't see square root

$\\ 2)\int \sqrt{\frac{x-1}{x+1}}dx \\=\int \frac{x-1}{\sqrt {(x-1)(x+1)}}dx \\=\int \frac{x-1}{\sqrt {x^2 -1}}dx \\=\int \frac{x}{\sqrt {x^2 -1}}dx - \int \frac{1}{\sqrt {x^2 -1}}dx \\=\frac{1}{2}\int (x^2-1)^{-\frac{1}{2}}\frac{d}{dx}(x^2-1)dx - ln( x + \sqrt {x^2 -1}) \\= \sqrt {x^2 -1} - ln( x + \sqrt {x^2 -1}) + C$

gurmies · Apr 10, 2009

Dang, beat me to it. Although the second question has a whole square root involved.

coeyz · Apr 10, 2009

thanks a lott~~
love you guys <3

gurmies · Apr 10, 2009

$\int \sqrt{\frac{x-1}{x+1}}dx = \int \frac{\sqrt{x-1}}{\sqrt{x+1}}dx$

$= \int \frac{x-1}{\sqrt{x^{2}-1}}dx$

$= \int \frac{x}{\sqrt{x^{2}-1}}dx - \int \frac{1}{\sqrt{x^{2}-1}}dx$

$= \sqrt{x^{2}-1} - ln (x + \sqrt{x^{2}-1})+ C$

coeyz · Apr 10, 2009

shinn said:
$1)\ \int \frac{1-2x}{x^2 + 2x + 3}dx \\= \int \frac{3}{x^2 + 2x + 3}dx -\int \frac{2x+2}{x^2 + 2x + 3}dx \\= 3\int \frac{1}{(x+1)^2+2}dx - ln (x^2 + 2x +3) \\= \frac{3 \sqrt{2}}{2}\tan^{-1} (\frac{x+1}{\sqrt{2}}) - ln (x^2 + 2x +3) + C$

edit: yea my bad, didn't see square root

$2)\int \sqrt{\frac{x-1}{x+1}}dx \\=\int \frac{x-1}{\sqrt {(x-1)(x+1)}}dx \\=\int \frac{x-1}{\sqrt {x^2 -1}}dx \\=\int \frac{x}{\sqrt {x^2 -1}}dx - \int \frac{1}{\sqrt {x^2 -1}}dx \\=\frac{1}{2}\int (x^2-1)^{-\frac{1}{2}}\frac{d}{dx}(x^2-1)dx - ln( x + \sqrt {x^2 -1}) \\= \sqrt {x^2 -1} - ln( x + \sqrt {x^2 -1}) + C$

for the first question, wht happen if the limit is 0 to 4?
coz when u sub it in i cant get the answer ;(
the answer is -4 square root 3 + 3 ln ( 5+3sqrt3 / 1+3sqrt3)
thanks againn

coeyz · Apr 10, 2009

coeyz said:
for the first question, wht happen if the limit is 0 to 4?
coz when u sub it in i cant get the answer ;(
the answer is -4 square root 3 + 3 ln ( 5+3sqrt3 / 1+3sqrt3)
thanks againn

oh can do it now~
thankss

gurmies · Apr 10, 2009

=) woot

coeyz · Apr 10, 2009

coeyz said:
oh can do it now~
thankss

oh no cant do it

i do it up to:

3/sqrt2 [ tan^-1 5/sqrt2 - tan^-1 1/sqrt2] - ln9

anyone please help :'(

gurmies · Apr 10, 2009

you sure that's the answer? I get ln (1/9) ...

coeyz · Apr 10, 2009

gurmies said:
you sure that's the answer? I get ln (1/9) ...

ye the ans is -4 square root 3 + 3 ln ( 5+3sqrt3 / 1+3sqrt3)

Drongoski · Apr 10, 2009

Could the answer be ??

$\frac{3\sqrt{2}}{2}tan^{-1}\frac{4\sqrt{2}}{7}\ \ -\ \ ln\ 9\ \ \ \ ?\ ?\ ?\ ?$

which is what you worked out, coeyz

coeyz · Apr 10, 2009

Drongoski said:
Could the answer be ??

$\frac{3\sqrt{2}}{2}tan^{-1}\frac{4\sqrt{2}}{7}\ \ -\ \ ln\ 9\ \ \ \ ?\ ?\ ?\ ?$

which is what you worked out, coeyz

okayy i think the answer is wrong dw
thanks a lot anyway

coeyz · Apr 10, 2009

may i also ask, how can u know that integrate x / sqrt (1-x^2) is - squrt 1-x^2 so quickly ??

coeyz · Apr 10, 2009

Drongoski said:
Could the answer be ??

$\frac{3\sqrt{2}}{2}tan^{-1}\frac{4\sqrt{2}}{7}\ \ -\ \ ln\ 9\ \ \ \ ?\ ?\ ?\ ?$

which is what you worked out, coeyz

how did u get the middle part?
what is [ tan^-1 (5/sqrt2) - tan^-1 (1/sqrt2) ] ?

GUSSSSSSSSSSSSS · Apr 10, 2009

coeyz said:
may i also ask, how can u know that integrate x / sqrt (1-x^2) is - squrt 1-x^2 so quickly ??

its just one where u have the derivative on top and the function on the bottom

and then the 2 and a half cancel out in the process

....lol i explained that real bad...sorry!

Drongoski · Apr 10, 2009

coeyz said:
how did u get the middle part?
what is [ tan^-1 (5/sqrt2) - tan^-1 (1/sqrt2) ] ?

Let phi = tan^-1 (5/sqr(2) ) - tan^-1(1/sqr(2))

Therefore: tan(phi) = [tan (tan^-1(5/(sqr2) - tan(tan^-1(1/sqr2)] / [ 1 + tan (tan^-1(5/sqr(2)) ) x tan(tan^-1(1/sqr 2))]

= [5/sqr 2 - 1 / sqr 2] / [ 1 + 5/sqr 2 x 1/sqr 2 ]

= 4/sqr 2 / (7/2) = 4 sqr 2 / 7

Therefore phi = tan^-1 (4 sqr 2)/7

gutzeit · Apr 11, 2009

I integrated terry lee until he un-permed his hair

gurmies · Apr 12, 2009

^ lolwut

gutzeit · Apr 17, 2009

gurmies said:
^ lolwut

LOL! Terry Lee also happens to be a guy in coeyz and my grade. Who unpermed his hair. He looked so cool before!

Integration question from terry lee :( (1 Viewer)

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