Find the integral of x/ (x^4+x^2+1) from 2 to 3.
I was going to integrate using the method for partial fractions but I can't factor x^4+x^2+1.
integrate x/(x^4+x^2+1)
| = integrate symbol
u = x^2
du = 2x dx
1/2 du = x dx
1/2 | du/(u^2 + u + 1)
Complete the square
u^2 + u = -1
u^2 + u + 1/4 = -1 + 1/4
(u + 1/2)^2 + 3/4
1/2 | du/((u + 1/2)^2 + 3/4)
w = u + 1/2
dw = du
1/2 | dw/(w^2 + 3/4)
1/2 | dw/(3/4 + w^2)
And, since,
| dx/(a^2 + x^2)= 1/a arctan x/a + C
the integration can easily be done now.
Can you take it from here?
Okay, so I ended up with the expression:
1/2[4/3arctan(4(x^2+1/2)/3)
Can you confirm this is correct before I use the fundamental theorem of calculus to determine the definite integral please.
1/2[2/(sqrt3)arctan(2(x^2+1/2)/(sqrt3)
I distributed the 1/2, but backed it out for my above answer...I don't think I made any errors backing it out.
Here is the answer, with the 1/2 distributed. I know this answer is right because I double check it with an online integration calculator.
1/(sqrt3)arctan (2x^2 + 1)/(sqrt3)
I didn't rationalize the denominators, because the online answer didn't.
Do you want me to post the final answer for you to check the definite integral?
That would be great if you could!
I'm working on getting the correct indefinite as you have above, but when I get it it would be great to have something to check my definite integral with :)
1/(sqrt3)arctan(5/(29(sqrt3))= 0.0572826
Check you values for 1/a and x/a
maybe that's where you got hung up
| dx/(a^2 + x^2)
(3/4 + w^2)
a^2 = 3/4
a = (sqrt3)/2
x^2 = w^2
x = w
So, you should have got,
1/2(1/(sqrt3/2)arctan (w/(sqrt3/2))
which simplified is,
1/2(2/(sqrt3)arctan(2w/(sqrt3))
then I distributed the 1/2
2/(2(sqrt3))arctan(2w/(sqrt3))
1/(sqrt3)arctan(2w/(sqrt3))
then substitute back in for u and x
Good Luck!!
To find the integral of x/(x^4 + x^2 + 1) from 2 to 3, we need to follow a different approach since factoring x^4 + x^2 + 1 into linear factors is not possible.
Instead, we can make use of a trigonometric substitution. Let's set x = tan(theta). By doing this substitution, we can rewrite the integral in terms of theta.
First, let's find the derivative of x with respect to theta using the chain rule:
dx/dθ = sec^2(θ)
Next, we need to express the expression x^4 + x^2 + 1 in terms of θ.
Since x = tan(θ), we can rewrite it as:
x^2 = tan^2(θ)
Now, let's substitute these expressions back into the integral:
∫[x/(x^4 + x^2 + 1)] dx
= ∫[tan(θ)/((tan^4(θ) + tan^2(θ) + 1)(sec^2(θ))] * sec^2(θ) dθ
= ∫[tan(θ)/((tan^4(θ) + tan^2(θ) + 1))] dθ
Now that we have completely transformed the integral into an integral involving θ, we can proceed to solve it.
However, trigonometric integrals involving tangent functions can be quite complex to evaluate unless the trigonometric identity is recognized. We can make use of the identity:
1 + tan^2(θ) = sec^2(θ)
Using this identity, we can rewrite the integral as follows:
= ∫[tan(θ)/(sec^4(θ) + sec^2(θ))] dθ
= ∫[tan(θ)/sec^2(θ)(sec^2(θ) + 1)] dθ
= ∫[sin(θ)/[(sec^2(θ))^2 + sec^2(θ)]] dθ
Let's simplify further using another trigonometric identity:
sin^2(θ) = 1 - cos^2(θ)
= ∫[sqrt(1 - cos^2(θ))/[(sec^2(θ))^2 + sec^2(θ)]] dθ
The integral now involves only cosine, which is easier to deal with. Let's make another substitution using the identity:
sin^2(θ) = 1 - cos^2(θ)
Let u = cos(θ), then du = -sin(θ) dθ
And when θ = 0, u = cos(0) = 1
And when θ = π/4, u = cos(π/4) = sqrt(2)/2
We can express the integral solely in terms of u:
= ∫[-sqrt(1 - u^2)/[(sec^2(u))^2 + sec^2(u)]] du
= -∫[sqrt(1 - u^2)/(sec^4(u) + sec^2(u))] du
Now, we can evaluate the integral of the above expression using standard integration techniques. The bounds of integration will also change to reflect the new variable u instead of θ.
After evaluating the definite integral, we can substitute back u = cos(θ) and find the final result.
Please note that trigonometric substitutions can sometimes yield complicated results, and it may be more efficient to use numerical methods or specialized software to obtain an accurate value for the definite integral.