Question

What is the antiderivative for 1/x^2-2x+2 ?

Answer 1

Just for it you have to that Integrate 1/x^2-2x+2.

Answer 2

How do I do it?

Answer 3

@abb0t Plz Help Us.

Answer 4

To integrate: \(\int\limits a^ndx= \frac{ a^{n+1} }{ n+1 }+C\) where C is some arbitrary constant. Also, NOTE: \[\frac{ 1 }{ a^n } = 1a^{-n}\]

Answer 5

@Dharmaputra_993 (x-1)² = x²-2x+1

Answer 6

1/(x^2-2x+2)

= 1/((x - 1)^2 + 1), completing the square. Let x - 1 = tan(u), so that

1/((x - 1)^2 + 1) = 1/(tan^2(u) + 1) = 1/sec^2(u); and

d(x - 1) = d(tan(u))

dx = sec^2(u)du; so the antiderivativ goes to

integral(1/((x - 1)^2 + 1))dx

= integral(sec^2(u)/sec^2(u))du

= integral(du) = u + C and since x - 1 = tan(u), u = arctan(x - 1), so this is just

arctan(x - 1) + C.

Answer 7

Right, sorry.

Answer 8

Oh, oops. Yeah, I think you want to use partial fractions to solve this.

Answer 9

Xxnina's approach is better.

Answer 10

I misread the question, sorry about that.

Answer 11

got it thank you (:

Answer 12

x^2-2x+2= (x-1)^2+1

so antiderivate of 1/x^2-2x+2
= antiderivate of 1/((x-1)^2+1)

= arctan (x-1) + C

Good Luck! I hope you understand!

Answer 13

x² - 2x + 2 = (x-1)² + 1

Answer 14

yeah i was stuck at how to approach it thank you so much. Nina did x^2-2x+2+1i believe.

Answer 15

I think it's -arctan

Answer 16

that would make it (1-x)

Answer 17

yes! great work! :)

Answer 18

(1-x) is the correct form of writing it usually. As you get higher in math, you'll notice that it's often writen as (1-x) rather than (-x+1) or (x-1).
It has to do with series.