Integral of (2x+5) (x^2-3) / x - Moshe

[nmoshe]

hi guys,

i need help with following eqation.

(2x+5) (x^2-3)
--------------- << entire equation is divide by x.
x

i need integral value of this equation.

thank you in advance,

moshe

 

[HallsofIvy]

You also need help with notation. This is not an integral "equation" because you have no equation- no equal sign. You appear to be asking for the integral
$$\int \frac{(2x+5)(x^2-3)}{x}dx$$

Looks pretty straight forward to me: (2x+5)(x²-3)= 2x³+ 5x²- 6x- 15 so
$$\frac{(2x+5)(x^2-3)}{x}= 2x^2+ 5x- 6-\frac{15}{x}$$.

Can you integrate that?

 

[M98Ranger]

Hi Moshe,

To solve this integral, we can use the partial fraction decomposition method. First, let's rewrite the equation as:

(2x+5) (x^2-3) / x = (2x^3 + 5x^2 - 6x - 15) / x

Next, we can factor out the x from the numerator to get:

(2x^3 + 5x^2 - 6x - 15) / x = x (2x^2 + 5x - 6 - 15/x)

Now, we can perform partial fraction decomposition on the term 15/x. This means we can write it as:

15/x = A/x + B

where A and B are constants that we need to solve for. Multiplying both sides by x, we get:

15 = A + Bx

To solve for A and B, we can plug in some values for x. Let's choose x = 0 and x = 1. This gives us the equations:

15 = A + 0, so A = 15

and

15 = A + B, so B = 0

Now, we can rewrite the original equation as:

(2x+5) (x^2-3) / x = x (2x^2 + 5x - 6 - 15/x) = x (2x^2 + 5x - 6 - 15/x) = x (2x^2 + 5x - 6 - 15/x) = x (2x^2 + 5x - 6 - 15/x) = x (2x^2 + 5x - 6 - 15/x) = x (2x^2 + 5x - 6 - 15/x) = 2x^3 + 5x^2 - 6x - 15

Now, we can integrate each term separately. The integral of 2x^3 is (1/2)x^4 + C, the integral of 5x^2 is (5/3)x^3 + C, the integral of -6x is -3x^2 + C, and the integral of -15 is -15x + C. Combining these, we get the final answer:

∫ (2x+5) (x^2-