Integrate: dx/(2x^2 + 4x + 7)
messy, are you sure you typed correctly?
I was hoping the bottom would factor so I could use partial fractions.
the way it stands I ran it through an integration program to get
(√10/10)[tan -1 [√10(x+1)/5]
are you working at that sophisticated level of integration??
Is this highschool level ??
You can just write the denominator as:
2x^2 + 4x + 7 =
2 (x^2 + 2 x + 7/2) =
2 [(x + 1)^2 + 5/2]
Then you use the fact that the integral of dx/(x^2 + a^2) = 1/a arctan (x/a)
So, the integral is:
1/2 1/sqrt(5/2) arctan[(x+1)/sqrt(5/2)]=
Answer given by Reiny above.
Thanks Reiny + Iblis!
This is from Wiley textbook "Calculus: Early Transcendentals Combined, 8th Edition", section 8.4. I think problem #41 (from memory).
I typed it right. The answer you two wrote matches the book, however I couldn't figure out how to do it. Makes sense now. Thanks
Reiny, thanks for writing that. I feel like an idiot getting stuck on these textbook problems sometimes.
To integrate the given expression, we can start by completing the square in the denominator.
Let's rewrite the denominator: 2x^2 + 4x + 7
First, divide both sides of the equation by 2 to simplify: x^2 + 2x + 7/2
To complete the square, we need to add and subtract (b/2)^2 where b is the coefficient of the x term. In this case, b = 2, so we add and subtract (2/2)^2 = 1.
So, adding and subtracting 1 to the expression: x^2 + 2x + 7/2 + 1 - 1
Now, let's group the terms: (x^2 + 2x + 1) + (7/2 - 1)
The first group can be written as a perfect square: (x + 1)^2
The second group simplifies: 7/2 - 1 = 7/2 - 2/2 = 5/2
Now, let's rewrite the integral:
∫ dx / (2x^2 + 4x + 7) = ∫ dx / ((x + 1)^2 + 5/2)
Since (x + 1)^2 appears in the denominator, we can use a trigonometric substitution to solve the integral.
Let's substitute: x + 1 = √(5/2) * tan(θ)
Taking the derivative of both sides with respect to x, we get: dx = √(5/2) * sec^2(θ) dθ
With these substitutions, the integral becomes:
∫ [√(5/2) * sec^2(θ) dθ] / (√(5/2) * tan^2(θ) + 5/2)
Simplifying, we get:
∫ sec^2(θ) / tan^2(θ) + 1 dθ
Using the identity: sec^2(θ) = tan^2(θ) + 1
The integral becomes:
∫ dθ
Finally, integrating dθ, we get:
θ + C
Replacing θ with the inverse tangent of (x + 1) / √(5/2):
∫ dx / (2x^2 + 4x + 7) = tan^(-1)((x + 1) / √(5/2)) + C
Therefore, the integral of dx / (2x^2 + 4x + 7) is tan^(-1)((x + 1) / √(5/2)) + C, where C is the constant of integration.