Indefinite Integrals Homework Help Needed

[Illusionist]

Homework Statement

Hi, I'm having trouble finding the indefinite integral of an equation.

Homework Equations

The function is (2x-13)/[(x^2-2x+4)^0.5]

The Attempt at a Solution

I thought it was a good idea to let u=x^2-2x+4, hence du/dx= 2x-2.
From that I got [-11/(u^0.5)].du but evaluating this gives me -22(u^0.5), which I don't believe is right.
Did I make a silly mistake or is my approach completely wrong? I am also having similar trouble finding the integral of [1+(1/t)]^5/(t^2).

Any help or advice would be very much appreciated, thank you.

 

[neutrino]

Your approach is correct, but you need rewrite the integral slightly.

$$\int{\frac{2x-13}{\sqrt{x^2-2x+4}}}$$

$$\int{\frac{2x-2}{\sqrt{x^2-2x+4}}} + \int{\frac{-11dx}{\sqrt{x^2-2x+4}}}$$

Now substitute $u=x^2-2x+4$, in the first integral.



As for the other integral, you can rewrite 1 + (1/t) as (t+1)/t.

 

[Qyzren]

for your second q, i would manipulate it to look like (t+1)^5/t^7 and use binominal expansion for the top.

 

[HallsofIvy]

$$\int \frac{2x-13}{\sqrt{x^2- 2x+ 4}} dx= \int \frac{2x-2}{\sqrt{x^2- 2x+4}} dx- \int \frac{11}{\sqrt{x^2-2x+ 4}}dx$$
Since the derivative of x²- 2x+ 4 is 2x- 2, the first can be done with the substitution u= x²-2x+ 4, but the second cannot.

Rewrite x²- 2x+ 4 as x²-2x+ 1+ 3= (x-1)²+3 (in other words, complete the square), let $v= \sqrt{3}(x-1)$ and then use a trig substitution

 

[neutrino]

and then use a trig substitution

Or as dexter would say, use a hyperbolic trig substitution.

 

[Illusionist]

Thank you soo much for all the help guys.
One question about HallsofIvy's comment. I understand why and how you completed the square. But don't understand why v=sq.rt.3(x-1). Now dv/dx=sq.rt.3. If I substitute this into the denominator I get a function that I don't know what to do with.
I tried to use standard integrals and basically pulled the 11 in front of the integration sign and was left with 1/[(x-1)^2+(3^0.5)^2]^0.5. Which using standard integral formula I was left with arcsinh[(x-1)/(3^0.5)]+c.
Again I suspect I am wrong with this approach.

 

[neutrino]

(-11)arcsinh[(x-1)/(3^0.5)]+c is correct.

 

[Illusionist]

Great. Thanks again everyone.