how do i integrate
2u du /u-2+2u^2
so i used subsitution rule?
or can i used integration by parts and how? thanks in adnvace
Is all of u -2 +2u^2
a denominator?
That is (2u-1)(u+1)
So you would be integrating
2u/[(2u-1)(u+1)]
The method of partial fractions can be used.
See
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html
To integrate the expression ∫(2u du)/(u-2+2u^2), you can indeed use the substitution method or integration by parts. Let's explore both methods step by step.
Method 1: Substitution
1. Let's start by letting w = u - 2 + 2u^2. Then, differentiate both sides of this equation with respect to u, which gives dw = (2 + 4u)du.
2. Rearrange the equation to solve for du: du = dw / (2 + 4u).
3. Substitute the new variables into the original integral: ∫(2u du)/(u-2+2u^2) = ∫(2u(dw / (2 + 4u)))/(w).
4. Simplify the integration: ∫(2u dw) / (w(2 + 4u)).
5. Factor out 2u in the numerator and cancel the common factors in the denominator: ∫(dw) / (w).
6. Integrate simply: ∫(1 / w) dw = ln|w| + C.
7. Bring back the original variable u: ln|w| + C = ln|u - 2 + 2u^2| + C.
Method 2: Integration by Parts
1. Apply the integration by parts formula: ∫ u dv = uv - ∫ v du.
2. Choose u and dv: Let u = 1, and dv = (2u du)/(u-2+2u^2).
3. Differentiate u to get du: du = 0 and integrate dv to get ∫(2u du)/(u-2+2u^2).
4. Apply the integration by parts formula: ∫(2u du)/(u-2+2u^2) = u∫dv - ∫vdu.
5. Evaluate u∫dv: Since u = 1 and ∫dv = ∫(2u du)/(u-2+2u^2), we get u∫dv = 1∫(2u du)/(u-2+2u^2) = ∫(2u du)/(u-2+2u^2).
6. Choose v and du: Let v = ln|u-2+2u^2|.
7. Integrate dv to get v: ∫v du = ∫ln|u-2+2u^2| du. This integral is non-elementary and difficult to solve analytically.
8. Combine the results from steps 5 and 7: u∫dv - ∫v du = ∫(2u du)/(u-2+2u^2) - ∫ln|u-2+2u^2| du.
As you can see, using the substitution method is simpler and more straightforward in this case because integration by parts leads to a non-elementary integral. Therefore, the final answer using the substitution method is ln|u - 2 + 2u^2| + C.