How do you find the integral of (x-1)/(x^2-4x+5)?

[Potatochip911]

Homework Statement

Find the integral of (x-1)/(x^2-4x+5)

Homework Equations

3. The Attempt at a Solution [/B]
After completing the square I get
∫(x-1)/((x-2)^2+1)*dx
Using substitutions x-2=tanθ; x-1=tanθ+1; dx=sec^2θ*dθ
∫[(tanθ+1)*(sec^2θ*dθ)]/(tan^2θ+1)
After cancelling
∫(tanθ+1)*dθ
integrating this I get
-ln(cosθ)+θ+C
ln(secθ)+θ+C
After substituting x back in
ln(x^2-4x+5)+arctan(x-2)+C
My answer key has a 1/2 infront of the natural log and I'm at a loss as to how they get it.

 

[phyzguy]

I think if you do the substitution properly, you will see that x^2-4x+5 = sec^2(θ), not sec(θ).

 

[Potatochip911]

I think if you do the substitution properly, you will see that x^2-4x+5 = sec^2(θ), not sec(θ).

I'm confused because with my substitution x^2-4x+5 turns into sec^2(θ) after using the identity tan^2(θ)+1=sec^2(θ)

 

[Mark44]

I think you misunderstood what phyzguy was saying. Here's what you wrote.

ln(secθ)+θ+C
After substituting x back in
ln(x^2-4x+5)+arctan(x-2)+C
My answer key has a 1/2 infront of the natural log and I'm at a loss as to how they get it.

Your substitution in the 3rd line above is wrong. sec(θ) = $\sqrt{x^2-4x+5}$, not the $x^2-4x+5$ that you have. The 1/2 that you should be seeing comes from this square root.

Also, I would have done a substitution with u = x - 2 before doing the trig substitution. Doing that makes life simpler, IMO.

 

[Potatochip911]

I think you misunderstood what phyzguy was saying. Here's what you wrote.

Your substitution in the 3rd line above is wrong. sec(θ) = $\sqrt{x^2-4x+5}$, not the $x^2-4x+5$ that you have. The 1/2 that you should be seeing comes from this square root.

Also, I would have done a substitution with u = x - 2 before doing the trig substitution. Doing that makes life simpler, IMO.

Whoops, yea I completely misunderstood and I will definitely try doing these with a u substitution for the x-a portion. Thanks!

 

[PeroK]

Whoops, yea I completely misunderstood and I will definitely try doing these with a u substitution for the x-a portion. Thanks!

Another approach is to split the integral up:

$\frac{x-1}{x^2-4x+5} = \frac{1}{2}\cdot \frac{2x - 4}{x^2-4x+5} + \frac{1}{x^2-4x+5}$

Then $\frac{1}{2}ln(x^2-4x+5)$ comes easily out of the first term as you have an exact derivative in the numerator.

 

[Potatochip911]

Another approach is to split the integral up:

$\frac{x-1}{x^2-4x+5} = \frac{1}{2}\cdot \frac{2x - 4}{x^2-4x+5} + \frac{1}{x^2-4x+5}$

Then $\frac{1}{2}ln(x^2-4x+5)$ comes easily out of the first term as you have an exact derivative in the numerator.

Yea this is how wolfram alpha solved it but I figured it would be a good idea to understand both methods incase it's useful later on

 

[HallsofIvy]

Equivalently, completing the square, $$x^2- 4x+ 5= x^2- 4x+ 4+ 1= (x- 2)^2+ 1$$. Let y= x- 2 so that dy= dx and y+ 1= x- 1.
So $$\int \frac{x- 1}{x^2- 4x+ 5} dx= \int \frac{y+ 1}{y^2+ 1} dy= \int \frac{y}{y^2+ 1} dy+ \int\frac{1}{y^2+ 1} dy$$

 

[Potatochip911]

Equivalently, completing the square, $$x^2- 4x+ 5= x^2- 4x+ 4+ 1= (x- 2)^2+ 1$$. Let y= x- 2 so that dy= dx and y+ 1= x- 1.
So $$\int \frac{x- 1}{x^2- 4x+ 5} dx= \int \frac{y+ 1}{y^2+ 1} dy= \int \frac{y}{y^2+ 1} dy+ \int\frac{1}{y^2+ 1} dy$$

Is it unusual that this integral can be solved in so many different ways?

 

[Mark44]

Is it unusual that this integral can be solved in so many different ways?

No, it happens fairly often. It's always true, though, that when you have two or more ways to evaluate an indefinite integral, the answers will differ by at most a constant. An example is $\int sin(x)cos(x)dx$. One person can use a substitution u = sin(x), du = cos(x)dx. Another person can use a different substitution, u = cos(x), du = -sin(x)dx. When they integrate, they get what appear to be different answers, namely (1/2) sin²(x) + C and (-1/2)cos²(x) + C. However, since sin²(x) = 1 - cos²(x), it turns out that the two answers are different only by a constant.

 

[Potatochip911]

No, it happens fairly often. It's always true, though, that when you have two or more ways to evaluate an indefinite integral, the answers will differ by at most a constant. An example is $\int sin(x)cos(x)dx$. One person can use a substitution u = sin(x), du = cos(x)dx. Another person can use a different substitution, u = cos(x), du = -sin(x)dx. When they integrate, they get what appear to be different answers, namely (1/2) sin²(x) + C and (-1/2)cos²(x) + C. However, since sin²(x) = 1 - cos²(x), it turns out that the two answers are different only by a constant.

Okay thanks that makes sense.