Even more intergration problems! (1 Viewer)

[Maianbarian]

Hey, we've just started integration and I'm struggling a bit with some of the questions, could someone show me how to do:

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

2. int ((4/(x^2-2x-1))dx

Thanks heaps

 

[Teoh]

1.
Get f'x on top (2x) so you can use the {f'x/fx = lnfx rule
So it ends up being
1/2 ln (x^2 + 1)

2.
Use 4(x-1)^-2
Then integrate from there

Disclaimer: Teoh often fails 4u maths assessments...

 

[Maianbarian]

Thanks teoh. I just did the first one it actually ends up being:
1/2ln(x^2+1)-arctanx
because you multiply the integrand by 2/2

With the second one that's what I did but I ended up with a mess so could somebody actually do the working for me? ( I assumed you meant 4/(x-1)^2 - 2 btw, otherwise you are doing something unknown to me)

 

[Constip8edSkunk]

1) yep, split up the fraction and the 1st part being what Teoh said, the remaining part, -1/x^2+1 integrates to -arctan x

2) the bottom equals (x-1-sqrt2)(x-1+sqrt2)
(just difference of squares)
split into partial fractions u get
sqrt2/(x-1-sqrt2) - sqrt2/(x-1+sqrt2)
(edit: you can skip most of the messy working by noticing that the x-1 part will cancel off while the sqrt2 will double if the 2 fractions are opposite signs => numerator *sqrt2 =2)
integrate to sqrt2 ln|(x-1-sqrt2)/(x-1+sqrt2)|

 

[Maianbarian]

ahhh, so that's how it works thankyou very much all is good now

 

[Maianbarian]

Ok, I've changed my mind all is not good... could you give me a hand with:

int (sqrt(16-x^2)/x^2) dx Hint: let x=4sinθ

I can get to:

=-(1/16)cotθ

but from there I've no idea

 

[Constip8edSkunk]

∫ sqrt(16 - x²)/x² dx
{x = 4sinθ
dx = 4cosθ dθ}
=∫ 4cosθ/16sin²θ 4cosθdθ
=∫ cot²θ dθ
=∫ cosec²θ-1 dθ
=-cot θ - &theta + C
=-cot arcsin x/4 - arcsin x/4 +C
=-sqrt(16-x²)/x - arcsin x/4 +C

edit: thanks xayma.... thought it worked on netscape... but not on ie

edit2: corrected eror

 

[Xayma]

Constip8edSkunk: You need to put ; after those signs for them to work

 

[Maianbarian]

I think you've made a mistake with your substitution in this line:

=∫ 4cosθ/16cos²θ 4cosθ dθ

I think you have let x=4cosθ insted of 4sinθ...

 

[Constip8edSkunk]

no sqrt (16-16sin²) = 4 cos

 

[Maianbarian]

yeah, that's fine for the top of the fraction but you also have cos² on the bottom. How did you get that?

 

[Constip8edSkunk]

oops, my mistake, i stand corrected.

u get cot² instead

this is cosec² - 1
integrate to give -cot θ - θ

edit: see original post

 

[Maianbarian]

Thanks Constip8edSkunk I see where I made my error now, it was very stupid too, *sigh* always the way...

 

[CrashOveride]

I_n in terms of I_n-1:

I_n = ∫ dx / (x² + a²)ⁿ

 

[CrashOveride]

What i mean to say is expresses I_n in terms of I_n-1

 

[CM_Tutor]

Originally posted by CrashOveride
I_n in terms of I_n-1:

I_n = ∫ dx / (x² + a²)ⁿ

I_n = ∫ dx / (x² + a²)ⁿ, for integers n > 1
= x / (x² + a²)ⁿ - ∫ x d(x² + a²)^-n, on integrating by parts
= x / (x² + a²)ⁿ - ∫ x * -n(x² + a²)^-n-1 * 2x dx
= x / (x² + a²)ⁿ + 2n * ∫ x² / (x² + a²)ⁿ⁺¹ dx
= x / (x² + a²)ⁿ + 2n * ∫ [(x² + a²) - a²] / (x² + a²)ⁿ⁺¹ dx
= x / (x² + a²)ⁿ + 2n * ∫ [1 / (x² + a²)ⁿ] - a² / (x² + a²)ⁿ⁺¹ dx
= x / (x² + a²)ⁿ + 2n * I_n - 2na² * I_n+1

So, I_n+1 = x / 2na²(x² + a²)ⁿ + [(2n - 1) / 2na²] * I_n

It follows that I_n = x / 2a²(n - 1)(x² + a²)^n-1 + [(2n - 3) / 2a²(n - 1)] * I_n-1, for integers n > 0

 

[CrashOveride]

Thanks for your reply CM, but i already solved it

Here's one for you now though:

if U_n = -1 --> 0 ∫ x^n (1+x)^1/2 dx

Show U_n = [(-2n) / (2n +3) ] U_n-1

 

[CrashOveride]

i get this as one of my final lines:

U_n ={ [ 2(1+x)^(3/2) x^n] / [3+2n] } - { [ (2n)/(3_2n) ] U_(n-1) }

with limits -1 --> 0

How do i apply the limits now to the U_(n-1) part? But it already, by its definition, has those limits? So i'd be applying the limits on the limits? o.o

 

[CM_Tutor]

Originally posted by CrashOveride
i get this as one of my final lines:

U_n ={ [ 2(1+x)^(3/2) x^n] / [3+2n] } - { [ (2n)/(3_2n) ] U_(n-1) }

with limits -1 --> 0

How do i apply the limits now to the U_(n-1) part? But it already, by its definition, has those limits? So i'd be applying the limits on the limits? o.o

It looks to me that you have the answer - the first term should be evaluated between the kimits of -1 ---> 0, which makes it disappear as its a term in x(1 + x), and the limits are already included in the U_n-1 part of the second term, so it should be left alone.

In general with parts, you should evaluate the 'uv' term when it appears if you are dealing with a definite integral.

I can write out the steps if this isn't clear, if you need me to.

 

[CrashOveride]

Actually just for reference, could you write it up

I may have more questions to follow