Integration Q (1 Viewer)

[shaon0]

1) S (2x+3)/sqrt(4-x^2) dx
2) S (1-2x)/sqrt(x^2+2x+10) dx

I know that you have to split up the integrals in both questions into partial fractions but don't know how to equate with the square roots.
I've used trig. substitutions and got the answer but i want to do it using partial fractions.

 

[tacogym27101990]

1) S (2x+3)/sqrt(4-x^2) dx
2) S (1-2x)/sqrt(x^2+2x+10) dx

I know that you have to split up the integrals in both questions into partial fractions but don't know how to equate with the square roots.

partial fractions is not used here, at least in my solution
id split the numerator in both.
1) S (2x+3)/sqrt(4-x^2) dx
= S 3/sqrt(4-x^2) dx + S 2x/sqrt(4-x^2) dx
which are both simple integrals
= 3asin(x/2) - 2sqrt(4-x^2)
2) split the integral into
S 3/sqrt(x^2+2x+10) dx - S 2x+2/sqrt(x^2 + 2 +10) dx
and work from there

cant be bothered finishing 2)
but if it isn't done by tomorrow i will

 

[Trebla]

Partial fractions would be highly inefficient for these integrals. The quickest way is to use the reverse chain rule or just substitute u = expression inside the square root.

 

[shaon0]

Partial fractions would be highly inefficient for these integrals. The quickest way is to use the reverse chain rule or just substitute u = expression inside the square root.

Yeah i've done it by using x= 2sin(u) and then proving cos(arcsin(x/2)) = sqrt(1-(x/2)^2). Is that alrite?
Solution:
1) S (2x+3)/sqrt(4-x^2) dx
Let x=2sin(u)
S (2x+3)/sqrt(4-x^2) dx
= S [4sin(u)+3]/[2cos(u)] dx
du/dx = (1/sqrt(4-x^2)) => 2cos(u)
Thus; S [4sin(u)+3]/[2cos(u)] dx
= S 4sin(u)+3 du
= -4cos(u) + 3u +C
= 3(arcsin(x/2)) - 4cos(arcsin(x/2))+C
= 3arcsin(x/2) - 4sqrt(1-(x/2)^2)+C
= 3arcsin(x/2)-2sqrt(4-x^2)+C

Thanks for the help though.

 

[shaon0]

partial fractions is not used here, at least in my solution
id split the numerator in both.
1) S (2x+3)/sqrt(4-x^2) dx
= S 3/sqrt(4-x^2) dx + S 2x/sqrt(4-x^2) dx
which are both simple integrals
= 3asin(x/2) - 2sqrt(4-x^2)
2) split the integral into
S 3/sqrt(x^2+2x+10) dx - S 2x+2/sqrt(x^2 + 2 +10) dx
and work from there

cant be bothered finishing 2)
but if it isn't done by tomorrow i will

Oh shit. My way was the long way. lol.
At least I found another method to getting that integral

 

[shaon0]

Question:
1) Integrate sqrt((x+2)/(x-3)) dx

For the answer, i got:
0.5sqrt(x^2-x-6)+(5/2)ArcCosh((2x-1)/5)+C
But i have a weird feeling that this answer isn't correct.

2) Evaluate S[limits:1 to 0] dx/(e^x+1) by splitting the numerator.
I just don't get the question lol

 

[Trebla]

Question:
1) Integrate sqrt((x+2)/(x-3)) dx

For the answer, i got:
0.5sqrt(x^2-x-6)+(5/2)ArcCosh((2x-1)/5)+C
But i have a weird feeling that this answer isn't correct.

2) Evaluate S[limits:1 to 0] dx/(e^x+1) by splitting the numerator.
I just don't get the question lol

1) ∫ √[(x + 2) / (x - 3)] dx
= ∫ (x + 2) / √[(x + 2)(x - 3)] dx
= ∫ (x + 2) / √[(x² - x - 6)] dx
= (1/2) ∫ (2x - 1 + 5) / √[(x² - x - 6)] dx
= (1/2) ∫ {(2x - 1) / √[(x² - x - 6)] + 5 / √[(x² - x + 1/4 - 25/4)]} dx
= (1/2) ∫ {(2x - 1) / √[(x² - x - 6)] + 5 / √[(x - 1/2)² - 25/4]} dx
= (1/2) {2√(x² - x - 6) + 5ln[x - 1/2 + √(x² - x - 6)]} + c
= √(x² - x - 6) + (5/2)ln[x - 1/2 + √(x² - x - 6)]} + c


2)
∫₀¹ dx / (e^x + 1)
= ∫₀¹ (1 + e^x - e^x) / (e^x + 1)
= ∫₀¹ (1 - (e^x / (e^x + 1))
= [x - ln(e^x + 1)]₀¹
= 1 - ln (e + 1) + ln 2
= 1 + ln (2 / (e + 1))

 

[shaon0]

1) ∫ √[(x + 2) / (x - 3)] dx
= ∫ (x + 2) / √[(x + 2)(x - 3)] dx
= ∫ (x + 2) / √[(x² - x - 6)] dx
= (1/2) ∫ (2x - 1 + 5) / √[(x² - x - 6)] dx
= (1/2) ∫ {(2x - 1) / √[(x² - x - 6)] + 5 / √[(x² - x + 1/4 - 25/4)]} dx
= (1/2) ∫ {(2x - 1) / √[(x² - x - 6)] + 5 / √[(x - 1/2)² - 25/4]} dx
= (1/2) {2√(x² - x - 6) + 5sin^-1[(2x - 1) / 5]} + c
= √(x² - x - 6) + 10sin^-1[(2x - 1) / 5] + c


2)
∫₀¹ dx / (e^x + 1)
= ∫₀¹ (1 + e^x - e^x) / (e^x + 1)
= ∫₀¹ (1 - (e^x / (e^x + 1))
= [x - ln(e^x + 1)]₀¹
= 1 - ln (e + 1) + ln 2
= 1 + ln (2 / (e + 1))

Oh alrite, Now i know where i went wrong for question 1.
Does it require a lot of practice to do integration very well and quickly?

 

[jet]

Does it require a lot of practice to do integration very well and quickly?

Not really. You just have to be able to adapt to each question. You know which require trig substitutions and which need partial fractions or IBP.
Once you look over the past 8-9 years of papers you'll see a pattern start to emerge which makes it alot easier to know what to expect.

 

[shaon0]

Not really. You just have to be able to adapt to each question. You know which require trig substitutions and which need partial fractions or IBP.
Once you look over the past 8-9 years of papers you'll see a pattern start to emerge which makes it alot easier to know what to expect.

Thanks. I usually use Substitutions of any kind. They seem a lot quicker for harder substitutions. I know when to use IBP though but i don't really know when to use partial fractions.

 

[azureus88]

partial fractions r the easiest i reckon. you just use it when both numerator and denominator are polynomials and the bottom is factorisable. when the degree of the top is bigger than or equal to the bottom, you gotta do polynomial division before carrying out partial fractions.

 

[jet]

Yeah. You split the numerator when you can't factorise, and you use partial fractions when you can. That's generally how I did it.
Splitting the numerator was especially helpful in Mechanics.

 

[shaon0]

Yeah. You split the numerator when you can't factorise, and you use partial fractions when you can. That's generally how I did it.
Splitting the numerator was especially helpful in Mechanics.

Alrite cool. I should practice a little more on partial fractions.
btw, Are you doing 17 units for your HSC?

 

[jet]

Well...I already did 4 of them, so 13 really.
But yeah, it adds up to 17.

 

[shaon0]

Well...I already did 4 of them, so 13 really.
But yeah, it adds up to 17.

wow. That's a lot of work to do even though its over 2 years.

 

[azureus88]

= (1/2) ∫ {(2x - 1) / √[(x² - x - 6)] + 5 / √[(x - 1/2)² - 25/4]} dx
= (1/2) {2√(x² - x - 6) + 5sin^-1[(2x - 1) / 5]} + c

are you sure this is right? the inverse sine bit seems a little dodgy. wouldnt that be the answer if it was 5 / √[(25/4) - (x - 1/2)²]} instead of 5 / √[(x - 1/2)² - 25/4]}. or maybe im missing something. can you please explain that step.

 

[Trebla]

Woops my bad!!! Check my edit...