Question

Evaluate the integral

Answer 1

f\[\int\limits \frac{ 2x-2 }{ x^2-4x+20 }dx\]

Answer 2

should i complete the square first in the denominator ?

Answer 3

so
I get (x^2-4x+4) +16

Answer 4

then that becomes (x+2)^2 + 16

Answer 5

Umm i don't know i'm a 5th grader i'm just going in problems :P sorry

Answer 6

then i get

\[\int\limits \frac{ 2x-2 }{ (x+2)^2+16 }\]

Answer 7

then i do u sub so i make my u = x+2 and my du=dx ? Im confused from here on

Answer 8

x^2-4x+4 is actually (x-2)^2 not (x+2)^2

Answer 9

ok thanks!

Answer 10

\[\int\limits \frac{2x-2}{(x-2)^2+16} dx \\ \int\limits \frac{2x-2}{16 \cdot (\frac{1}{16}(x-2)^2+1)} \\ \frac{2}{16} \int\limits \frac{x-1}{(\frac{x-2}{4})^2+1} dx \\ \frac{1}{8} \int\limits \frac{x-1}{(\frac{x-2}{4})^2+1} dx\]

Answer 11

you can do a trig sub

Answer 12

why did you multiply the 16? i thought i was supposed to add it?

Answer 13

do you know the distributive property?

Answer 14

ab+ac=a(b+c)

Answer 15

ohh ok i see!

Answer 16

I was putting the bottom in the form that is familair such as the left hand side of
\[\tan^2(\theta)+1=\sec^2(\theta)\]

Answer 17

\[\tan(\theta)=\frac{x-2}{4}\]

Answer 18

i also don't understand the (x-2)/(4) in the third step..

Answer 19

thats equal to tan?

Answer 20

4^2 is 16

Answer 21

so will the next step sec^2 theta on the denominator ?

Answer 22

\[(x-2)^2+16 \\ \text{ so that this is easier for you to see } \\ \text{ I'm going to put a fancy 1 next to the } (x-2)^2 \\ \frac{16}{16}(x-2)^2+16 \\ \text{ now I'm going to factor out 16 } \\ 16 ( \frac{1}{16}(x-2)^2+1) \\ 16(\frac{(x-2)^2}{16}+1) \\ \text{ now recall } 4^2=16 \\ 16(\frac{(x-2)^2}{4^2}+1) \\ 16((\frac{x-2}{4})^2+1)\]

Answer 23

yes since tan^2(theta)+1 is sec^2(theta)

Answer 24

\[4 \tan(\theta)=x-2 \\ 4\tan(\theta)+1=x-1 \text{ added 1 on both sides }\]

so you can also replace x-1 in terms of theta now
you will also need to express dx in terms of theta and d theta

Answer 25

\[4 \tan(\theta)=x-2 \\ 4 \sec^2(\theta) d \theta= dx \]

Answer 26

\[\int\limits\limits \frac{2x-2}{(x-2)^2+16} dx \\ \int\limits\limits \frac{2x-2}{16 \cdot (\frac{1}{16}(x-2)^2+1)} \\ \frac{2}{16} \int\limits\limits \frac{x-1}{(\frac{x-2}{4})^2+1} dx \\ \frac{1}{8} \int\limits\limits \frac{x-1}{(\frac{x-2}{4})^2+1} dx \\ \frac{1}{8} \int\limits \frac{4 \tan(\theta)+1}{\tan^2(\theta)+1} 4 \sec^2(\theta) d \theta \]
and yes you can write that bottom there as sec^2(theta)

Answer 27

\[\int\limits\limits\limits \frac{2x-2}{(x-2)^2+16} dx \\ \int\limits\limits\limits \frac{2x-2}{16 \cdot (\frac{1}{16}(x-2)^2+1)} \\ \frac{2}{16} \int\limits\limits\limits \frac{x-1}{(\frac{x-2}{4})^2+1} dx \\ \frac{1}{8} \int\limits\limits\limits \frac{x-1}{(\frac{x-2}{4})^2+1} dx \\ \frac{1}{8} \int\limits\limits \frac{4 \tan(\theta)+1}{\tan^2(\theta)+1} 4 \sec^2(\theta) d \theta \\ \frac{1}{8} \int\limits \frac{4 \tan(\theta)+1}{\cancel{\sec^2(\theta)}} 4 \cancel {\sec^2(\theta)} d \theta \]
try to continue from here

Answer 28

thanks so much! i will follow what you did thanks

Answer 29

let me know if you have anymore questions on this one
I will be sorta around; looking at my computer every few minutes (multi-tasking a bit here)

Answer 30

ok great thanks!