how do u integrate 1+u/3u+2u^2
please help me get started ...tnk u
I would do partial fractions.
assume you mean
(1+u) du /(3u+2u^2)
which is
(1+u) du/[u(3+2u)]
which is
1 du/[u(3+2u)] + u du/[u(3+2u)]
now let's see if we can expand 1/[u(3+2u)]
into
a/u + b/(3+2u) = 1/ [u(3+2u)]
a(3+2u) + b(u) = 1
3 a = 1 so a = 1/3
2a + b = 0 so b = -2/3
so we have
1/[u(3+2u)] = (1/3)/u - (2/3)/(3+2u)
so the integral now is
(1/3)du/u -(2/3)du/(3+2du) +du/(3+2u)
or
(1/3)du/u +(1/3) du/(3+2u)
ok from there?
To integrate the expression 1 + u / (3u + 2u^2), you can use the technique of partial fractions decomposition. Here's how you can get started:
Step 1: Factorize the denominator
Start by factoring the denominator, which in this case is (3u + 2u^2). We can rewrite it as u(3 + 2u).
Step 2: Setting up the partial fractions decomposition
Since the numerator is degree 0 (a constant), and the denominator can be factored as a linear and quadratic expression, we can express the integrand as the sum of two partial fractions:
1 + u / (3u + 2u^2) = A / u + B / (3 + 2u)
Step 3: Finding the values of A and B
To find the values of A and B, we need to solve for them by equating the numerators:
1 + u = A(3 + 2u) + Bu
Now, we need to simplify and rearrange this equation to solve for A and B.
Step 4: Solving for A and B
First, expand the right-hand side of the equation:
1 + u = 3A + 2Au + Bu
Next, group the terms with the same power of u:
(2A + B)u + (3A - 1) = 0
Since both sides of the equation must be equal to zero, we can equate the coefficients of the u terms and the constant terms:
2A + B = 1 (Equation 1)
3A - 1 = 0 (Equation 2)
Now, solve these two equations simultaneously to find the values of A and B.
In Equation 2, rearrange to solve for A:
3A = 1
A = 1/3
Substitute the value of A back into Equation 1 to solve for B:
2(1/3) + B = 1
2/3 + B = 1
B = 1 - 2/3
B = 1/3
So, A = 1/3 and B = 1/3.
Step 5: Writing the integral as a sum of the partial fractions
Now that we have the values of A and B, we can rewrite the initial integral:
1 + u / (3u + 2u^2) = (1/3) / u + (1/3) / (3 + 2u)
Step 6: Integrating the partial fractions
Finally, integrate each of the partial fractions separately. In this case, the integral of 1/u is ln|u|, and the integral of 1/(3 + 2u) can be found using substitution or a known integral formula.
Integrating ln|u| gives you ln|u|, and integrating 1/(3 + 2u) gives you (1/2) ln|3 + 2u|.
Therefore, the final integral of the expression 1 + u / (3u + 2u^2) is:
∫ (1/3) / u du + ∫ (1/3) / (3 + 2u) du = (1/3) ln|u| + (1/6) ln|3 + 2u| + C
where C is the constant of integration.