How do you integrate #int (2x-2)/ (3 -2x - x^2)^(1/2)dx# using partial fractions?

Question

Evelyn Agard · Accepted Answer

See the answer below:

Ezra Adams · Answer

undefined To integrate $\int \frac{{2x - 2}}{{\sqrt{3 - 2x - x^2}}}dx$ using partial fractions, follow these steps:

1. Factor the denominator to identify its roots.
2. Decompose the fraction into partial fractions based on the factors of the denominator.
3. Solve for the coefficients of the partial fractions.
4. Integrate each partial fraction separately.
5. Combine the results to find the overall integral.

Given that the denominator $3 - 2x - x^2$ factors as $-(x + 1)(x - 3)$, the partial fraction decomposition is:

$$\frac{{2x - 2}}{{\sqrt{3 - 2x - x^2}}} = \frac{{A}}{{\sqrt{x + 1}}} + \frac{{B}}{{\sqrt{x - 3}}}$$

Now, solve for $A$ and $B$ by multiplying through by the common denominator:

$$2x - 2 = A\sqrt{x - 3} + B\sqrt{x + 1}$$

After finding $A$ and $B$, integrate each term separately:

$$\int \frac{{A}}{{\sqrt{x + 1}}}dx = A\sqrt{x + 1} + C_1$$
$$\int \frac{{B}}{{\sqrt{x - 3}}}dx = B\sqrt{x - 3} + C_2$$

Finally, combine the results and add the constant of integration to get the overall integral:

$$\int \frac{{2x - 2}}{{\sqrt{3 - 2x - x^2}}}dx = A\sqrt{x + 1} + B\sqrt{x - 3} + C$$