Indefinite Integrals Homework Help Needed

• Illusionist
In summary, the student is having trouble finding the indefinite integral of an equation and is also having trouble finding the indefinite integral of [1+(1/t)]^5/(t^2). His approach is correct, but he needs to rewrite the integrals slightly.f

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.

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.

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

$$\int \frac{2x-13}{\sqrt{x^2- 2x+ 4}} dx= \int \frac{2x-2}{\sqrt{x^2- 2x+4}} dx- \int \frac{11}{\sqrt{x^2-2x+ 4}}dx$$
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

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

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.

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

Great. Thanks again everyone.