Intergration Using Substitution (1 Viewer)

[FDownes]

I've got the basics of 'Intergration Using Substitution' down, but I'm having some trouble with the more complex questions. I was hoping someone might be able to go through one of the questions step-by-step so that I can understand it better.

So here's the question;

By using the substitution u^2 = x + 2, find ∫(x - 2)/[√(x + 2)] dx.

Can anyone help me out?

 

[A High Way Man]

OMG!!! this is the first time i can be the first to reply without someone else stealing my thunder.

well you have to use

u^2 = x + 2

differentiating you have

2u du = dx

so, thats easy enough

the integral is (x+2)dx/(x+2)^(1/2)

so, just substitute

the integral becomes
u (2u du)/u^2

= 2u^2 du

integrate that and you get

2u^3/3 as your indefinate integral

then sub u back in

[2(x+2)^3/2]/3 + C

 

[FDownes]

I made the same mistake the first time I tried to answer the question. The intergral isn't ∫(x + 2)/[√(x + 2)] dx, it's ∫(x - 2)/[√(x + 2)] dx.

Sorry.

 

[Shadose]

I got one I need help with too, can someone help me with this?

∫dx/(e^x+1) , using z=e^x+1

I would much appreciate it ^^

 

[Slidey]

Er ok...
z=e^x+1
z-1=e^x
dz=e^x dx

So to multiply by dz we also have to divide by e^x (since there's no e^x term present):

I(1/z * 1/e^x dz)
z-1=e^x, so:
I(1/z * 1/(z-1) dz)
Now use partial fractions:

A/z + B/(z-1) = 1/(z(z-1))
A(z-1)+Bz=1
let z=1, B=1
let z=0, A=-1, so:

I(1/z * 1/(z-1) dz) = I(1/(z-1) - 1/z dz)
= ln(z-1) - lnz + C

Now, z=e^x+1
z-1=e^x, so:

Answer is: ln(e^x) - ln(e^x+1) + C
= x - ln(e^x+1) + C

Test this answer by deriving:
Derivative is: 1 - e^x/(e^x+1) = (e^x+1)/(e^x+1) - e^x/(e^x+1) = 1/(e^x+1)

Yay. Although I used a 4unit method (partial fractions). Perhaps somebody can try this from a 3unit perspective.

 

[Slidey]

LOL. Examining the end of my post where I derived my answer to confirm it, a simpler method struck me:

1/(e^x+1) = (e^x+1)/(e^x+1) - e^x/(e^x+1) = 1 - e^x/(e^x+1), so:
z=e^x+1
dz=e^x dx

Thus: I dx - I e^x/(e^x+1) dx = x - I dz/z = x - lnz + C = x - ln(e^x+1) + C.

Done.

 

[Shadose]

LOL. Examining the end of my post where I derived my answer to confirm it, a simpler method struck me:

1/(e^x+1) = (e^x+1)/(e^x+1) - e^x/(e^x+1) = 1 - e^x/(e^x+1), so:
z=e^x+1
dz=e^x dx

Thus: I dx - I e^x/(e^x+1) dx = x - I dz/z = x - lnz + C = x - ln(e^x+1) + C.

Done.

Thnks ^^ Thats so smart!!

 

[Slidey]

Remember that trick. It's fairly important and has saved my arse numerous times in maths questions and exams!

E.g.: what's the asymptote of y=x/(x+1)?

y=(x+1-1)/(x+1) = 1 - 1/(x+1)

So the asymptote is y=1 as x-> infinity.