MHB How Do You Solve Integrals Using Partial Fractions?

karush
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partial fractions
$$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}
=\frac{A}{(x^2+5)}+\frac{B}{(x-3)}$$
$$3x^2+x+12=A(x-3)+B(x^2+5)$$
x=3 then 27+3+12=14B
3=B
x=0 then
12=-3A+15
1=A
$$\int\frac{1}{(x^2+5)} \, dx
+3\int\frac{1}{(x-3)}\, dx$$ $\displaystyle
\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}}
+3\ln\left(\left|x-3\right|\right)+C$

maybe? not sure
 
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karush said:
partial fractions
$$\int\frac{3x^2+x+12}{(x^2+5)(x-3)}
=\frac{A}{(x^2+5)}+\frac{B}{(x-3)}$$
$$3x^2+x+12=A(x-3)+B(x^2+5)$$
x=3 then 27+3+12=14B
3=B
x=0 then
12=-3A+15
1=A
$$\int\frac{1}{(x^2+5)} \, dx
+3\int\frac{1}{(x-3)}\, dx$$ $\displaystyle
\frac{\arctan\left(\frac{x}{\sqrt{5}}\right)}{\sqrt{5}}
+3\ln\left(\left|x-3\right|\right)+C$

maybe? not sure

Try differentiating your answer to see if you get back to where you started. You should quickly see that it is incorrect.

The degree of each numerator needs to be ONE DEGREE LESS than the degree of the denominator. So that means you need to have $\displaystyle \begin{align*} \frac{A\,s + B}{x^2 + 5} + \frac{C}{x - 3} \end{align*}$ as your partial fraction expansion.
 
Typo: Prove It meant \frac{Ax+ B}{x^2+ 5}+ \frac{C}{x- 3}.

Also, he said "one degree less than the denominator". Actually you can always factor a polynomial into linear or quadratic factors (with real coefficients) so you need linear numerators, (Ax+ B), over quadratic denominators and constants (C) over linear factors.

If we were willing to use complex numbers, we can write any polynomial as linear factors and we could write this is \frac{A}{x+ 5i}+ \frac{B}{x- 5i}+ \frac{C}{x- 3}. Of course we could then go back to real numbers by rationalizing the denominators: \frac{A}{x+ 5i}\frac{x- 5i}{x- 5i}+ \frac{B}{x- 5i}\frac{x+ 5i}{x+ 5i}= \frac{Ax+ 5iA+ Bx- 5Bi}{x^2+ 5}= \frac{(A+ B)x+ 5i(A- B)}{x^2+ 5}. Here, it will turn out that both A+ B is real and A- B is imaginary so that both A+ B and 5i(A- B) are real.
 
https://dl.orangedox.com/geAQogxCM0ZYQLaUju

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