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hello

i think that i need to get this in the form 1/u^2 +1 du so the integral is arctan(u) +c
could you please tell me how to do this

Also could you tell me when the answer to an integral is arcsin(u).
The arccos(u) i believe is 1/x^2 -1 du

thank you;-)

2007-02-17 02:35:23 · 8 answers · asked by mond257 1 in Science & Mathematics Mathematics

8 answers

Hi

here you have something of the form 1/u^2+0, so your answer won't be arctan(u).

What you need to do is to change variables.
Put x+2 = t
then differentiate both sides to obtain
dx=dt

This means you will replace (x+2) by t and dx by dt so your integral is now
∫ [1/t^2 dt]
= ∫ [t^(-2) dt]
= (-1) t^(-1) + C
= -1/t + C (now recall t=x+2)
= -1/(x+2) + C

Hope this helps.

2007-02-17 03:29:26 · answer #1 · answered by M 6 · 2 0

1/x^2+4

2007-02-17 06:53:51 · answer #2 · answered by Dead Dog 2 · 0 0

The integral of 1/(x+2)^2 is -1/(x+2) + C

To answer your further query,
1/(1-x^2)^1/2 has integral arcsin(x)
-1/(1-x^2)^1/2 has integral arccos(x)

2007-02-17 07:15:01 · answer #3 · answered by aepacino 2 · 0 0

let x+2=u
then differentiate this to get: du/dx=1
so du=dx
now, 1/(x+2)^2 dx = 1/(u)^2 du = u^-2 du = -u^-1 + c= -(x+2)^-1+c
where c is a constant of integration
hope u understand now!!

2007-02-17 14:07:59 · answer #4 · answered by kate 1 · 0 0

int(1/(x+2)^2)dx

put u=(x+2)
>>>>du=dx

int(1/(x+2)^2)dx
=int(1/u^2)du
= -1/u+C
= -1/(x+2) +C
{substituting (x+2) back}

where C is a constant

now your second part
let y=arcsinx
x=siny
dx/dy=cosy>>dy/dx=1/cosy
=1/sqrt(1-sin^2) {pythagoras}
=1/sqrt(1-x^2)
therefore,
int(1/sqrt(1-x^2)dx
=arcsin x+C
{differentiation is the reverse
of integration}
it can be shown that
int(-1/sqrt(1-x^2)dx
=arcos(1-x^2)+C
by a similar argument

i hope that this helps

2007-02-17 07:47:31 · answer #5 · answered by Anonymous · 0 0

∫ 1/(x+2)^2 dx = - 1/(x+2) + C

because differentiating 1/(x+2) with respect to x yields -1/(x+2)^2.

2007-02-17 02:41:48 · answer #6 · answered by MHW 5 · 1 0

(one million - x)^4 = (one million - 2x + x²)² = one million - 2x + x² -2x + 4x² - 2x³ + x² - 2x³ + x^4 = one million - 4x + 6x² - 4x³ + x^4 Multiply that via x^4: x^8 - 4x^7 + 6x^6 - 4x^5 + x^4 using long branch: ......... x^6 - 4x^5 + 5x^4 - 4x^2 + 4 .........???????????????????? x² + one million|x^8 - 4x^7 + 6x^6 - 4x^5 + x^4 .........-x^8 + 0x^7 - x^6 .........???????????? ..................-4x^7 + 5x^6 - 4x^5 ...................4x^7 + 0x^6 + 4x^5 ...................???????????? ..............................5x^6 + 0x^5 + x^4 .............................-5x^6 + 0x^5 -5x^4 .............................??????????... .......................................... .......................................... + 4x^2 .......................................... .......................................... ........................................... - 4 .......................................... .......................................... The -4 the rest ability that: ((x^4)(one million-x)^4)/(one million+x^2) = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - 4/(x² + one million) The essential turns into 6 relatively straight forward integrals; all with greater limits = one million and decrease limits = 0: ?(x^6)dx - 4?(x^5)dx + 5?(x^4)dx - 4?(x^2)dx + 4?dx -4?(one million/(x² + one million))dx

2016-10-15 12:33:15 · answer #7 · answered by fugere 4 · 0 0

Read your text books !!

2007-02-17 02:38:47 · answer #8 · answered by Anonymous · 0 0

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