Integral of √(x-2)/(x+1) after u-substitution

[ISU20CpreE]

ok i got this problem:

$$\int\frac{\sqrt{x-2}} {x+1} \,dx$$

then i have to let u=(x-2)^0.5 then i solve for x in terms of u

so i get x= u^2+2
dx=2u du

$$\int\frac{\sqrt{x-2}} {x+1} \,dx=\int\frac{u}{u^2+3}\,2u\,du$$

where when i substitute i get:

$$\int\frac{2u^2} {u^2+3}\,du$$. after that I am completely lost.

 

[mattmns]

Maybe:

Pull the 2 out, then add 0 to the top (+3 - 3). Then, can you solve: $$2\int\frac{u^2 + 3 - 3} {u^2+3}\,du$$

 

[Cyrus]

Try making u = x+1

 

[ISU20CpreE]

Maybe:

Pull the 2 out, then add 0 to the top (+3 - 3). Then split the fractions and easy.

what if i do this
$$\frac{2u^2}{u^2+3}=2-\frac6{u^2+3}$$ i do polynomial long division. lol. but then what??

 

[ISU20CpreE]

Try making u = x+1

It will be easier but the book says i have to use the other one.

 

[mezarashi]

what if i do this
$$\frac{2u^2}{u^2+3}=2-\frac6{u^2+3}$$ i do polynomial long division. lol. but then what??

Then it's cool. The first time is easily integrable. The second term has a solution as well.

Edit: Integrating the second term will be easier with substitution $$u = a\tan y$$, and the identity $$\tan^2 x + 1 = \sec^2 x$$

 

[mattmns]

Did you see my edit? I think that is really easy to solve. No polynomial division needed

 

[Cyrus]

$$2\int\frac{u^2 + 3 - 3} {u^2+3}\,du$$


how is that helpful?

 

[mattmns]

$$2\int\frac{u^2 + 3 - 3} {u^2+3}\,du$$


how is that helpful?

Split that up into two parts:

(a+b)/ c = a/c + b/cSo,
$$2\int\frac{u^2 + 3 - 3} {u^2+3} du = 2(\int\1du - \int\frac{3}{u^2+3}du)$$

Which is easy to solve, unless I made some silly mistake.

 

[ISU20CpreE]

Maybe:

Pull the 2 out, then add 0 to the top (+3 - 3). Then, can you solve: $$2\int\frac{u^2 + 3 - 3} {u^2+3}\,du$$

I am confused with differnent methods sorry I've been trying for a few hours. I don't know what to do.

 

[Cyrus]

Gotcha mattmns, your right that is easier.


looking at my table of integrals:

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

where x is your u, and a is square root of 3.

 

[ISU20CpreE]

Gotcha mattmns, your right that is easier.


looking at my table of integrals:

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

so if i will get as an answer this??

$$2u-6\frac{1} {\sqrt{3}} tan^-1\frac{u} {\sqrt{3}},+c$$

 

[Cyrus]

change all your U's into $$\sqrt{x-2}$$, and you can anwser your own question, if we did this right, take the derivative, do you get the question you were asked in the beginning? That is always a good check to do when your not certain.

 

[ISU20CpreE]

change all your U's into $$\sqrt{x-2}$$, and you can anwser your own question, if we did this right, take the derivative, do you get the question you were asked in the beginning? That is always a good check to do when your not certain.

You are totally right my friend, but in the book it gives me this answer

$$2\sqrt{x-2}-2\sqrt{3}\tan^{-1}\left(\sqrt{\frac{x-2}3}\right)+C$$

 

[Cyrus]

hmmm the inner part of the ( ) looks good, some how I goofed off the 2 (3)^.5 part let me see.

nope were still fine, just multiply top and bottom by square root of three. so that you go from $$\frac { 6} { \sqrt 3 }$$

to

$$\frac { 6 \sqrt 3} { ( \sqrt 3)^2 }$$

and you get

$$2 \sqrt {3}$$

 

[ISU20CpreE]

hmmm the inner part of the ( ) looks good, some how I goofed off the 2 (3)^.5 part let me see.

nope were still fine, just multiply top and bottom by square root of three.

LoL, no problem thanks. Thank you very much for the help.

I see i really needed it