The following definite integral can be evaluated by subtracting F(B) - F(A), where F(B) and F(A) are found from substituting the limits of integration.

\int_{0}^{4} \frac{1600 x +1200 }{(2 x^2 +3 x +1)^5}dx

After substitution, the upper limit of integration (B) is :
and the lower limit of integration (A) is :

After integrating,

F(B) =

F(A) =

pretty easy upper limit of b is 25.15 lower limit of a is 16.23

udk

To find the upper and lower limits of integration, we look at the given definite integral \(\int_{0}^{4} \frac{1600 x +1200 }{(2 x^2 +3 x +1)^5}dx\). The upper limit of integration (B) is the value at which we evaluate the function, in this case, x = 4. The lower limit of integration (A) is the other value at which we evaluate the function, in this case, x = 0.

Now, let's evaluate the indefinite integral of the given function \(\frac{1600 x + 1200 }{(2 x^2 + 3 x + 1)^5}\) to find F(x).

To integrate this function, we can use a substitution method.

Let \(u = 2 x^2 + 3 x + 1\), then differentiating both sides with respect to x, we get \(\frac{du}{dx} = 4x + 3\).

Rearranging the terms, we have \(du = (4x + 3) dx\).

Now, we substitute the values in the integral as follows:

\(\int \frac{1600 x +1200 }{(2 x^2 +3 x +1)^5}dx = \int \frac{(1600 x + 1200)dx}{u^5}\)

Since \(du = (4x + 3) dx\), we can rewrite the integral as:

\(\int \frac{1600 x +1200 }{(2 x^2 +3 x +1)^5}dx = \int \frac{(1600 x + 1200)dx}{u^5} = \int \frac{(1600 x + 1200)dx}{u^5} = \int \frac{1}{u^5} du\).

Integrating \(\frac{1}{u^5}\) with respect to u gives us \(-\frac{1}{4u^4}\).

Therefore, we have:

F(x) = \(-\frac{1}{4(2 x^2 + 3 x + 1)^4} + C\),

where C is the constant of integration.

Next, we substitute the upper limit of integration (B = 4) and lower limit of integration (A = 0) into the function F(x) to find the values F(B) and F(A).

F(B) = \(-\frac{1}{4(2(4)^2 + 3(4) + 1)^4}\) + C

F(A) = \(-\frac{1}{4(2(0)^2 + 3(0) + 1)^4}\) + C