Integrating a fraction?

Hello.

Use partial fractions so you can integrate.

(3x^2 + 2x +4) / ((x+1)*(x^2 + 4))

3x^2 + 2x +4 = (A/x+1) + (Bx + C)/(x^2 + 4)

3x^2 + 2x +4 = A(x^2 + 4) + (Bx + C)*(x+1) ------ (3)

Set x = -1 to find out the value of A:

5 = A(1 +4) ----- 5 = 5A ------ A = 1

Now we equate the coefficients (multiply out in nr. 3)

3 = 1 + B ----- B = 2

Line 3 multiplied out looks like this:

Ax^2 + 4A + Bx^2 + Cx + C ------ set x = 0

4 = 4A + C --- A =1 so 4 + C = 4 means C = 0

That´s how we end up with what you stated in your question. You also say that integrating it is no deal so I assume you know how to do that.

Hope this helped!

Z = ∫ (3x² + 2x + 4) / ((x+1)*(x² + 4)) dx

Z = ∫ A / (x+1) dx + ∫ (Bx + C)/(x² + 4)) dx

Z = ∫ A(x² + 4) / [(x+1)(x² + 4)] dx + ∫ (Bx + C)(x + 1)/[(x+1)(x² + 4)] dx

do algebra operation just on numerator only,

for x = -1

A[(-1)² + 4] = 5A = 3(-1)² + 2(-1) + 4 = 5

A = 1

for x = 0

A(0² + 4) + (B(0) + C)(0 + 1) = 4(1) + C = 3(0)² + 2(0) + 4 = 4

C = 0

for x = 1

A[1² + 4] + (B(1) + C)(1 + 1) = 1(5) + 2B + 2(0) = 3(1)² + 2(1) + 4 = 9

B = 2

Z = ∫ 1 / (x+1) dx + ∫ 2x/(x² + 4)) dx

Z = ∫ d(x + 1) / (x+1) + ∫ d(x² + 4)/(x² + 4))

Z = ln |x + 1| + ln |x² + 4| + C

As you probably know, the integral in this type of question is simply ln((x+1)*(x^2 + 4))

or ln(x+1) + ln(x^2 + 4) if you prefer. There are only about five or six methods of integration, and you have to hunt around for the method that applies in any particular case. It's quite easy to write functions that can't be integrated !

f'(x) = 2/3 * x^4 dx f(x) = 1/5 * 2/3 * x^5 f(x) = 2/15 * x^5

Let u = (x+1)*(x²+4)

du/dx = (3x² + 3x + 4)

integral of du/dx/u du/udx = log((x+1)(x² + 4)) + C

We showed this to " Oiru " a few hours ago.