Computation Of Integrals

Question

Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$ Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator to get $$2x+1=A(x-2)+B(x-1)$$ Now to find $A$ and $B$ how can we put $x=1$ and $x=2$ in this identity as this identity is valid if and only if $x$ is not equal to $1, 2$

My question is, in many books this method is used but how is this possible — user679770, Aug 08 '19 at 17:38
Are you asking why the result works for $x=1,2$ also? Two linear polynomials that agree at two points agree everywhere. (Two points determine a line.) — saulspatz, Aug 08 '19 at 17:40
The fraction $$\dfrac{A}{x-1}$$ has no discontinuity at $x=2$. And similarly $$\dfrac{B}{x-2}$$ has no discontinuity at $x=1$. This implies that the numerator should be continuous, and therefore, we can plug in values of the discontinuities (that would make a denominator zero), and the values the constants take at those values should be the same as the values they would take at any other point. — SlipEternal, Aug 08 '19 at 17:43

John Omielan · Answer 1 · 2019-08-08T18:33:33.697

You have

$$2x+1=A(x-2)+B(x-1) \tag{1}\label{eq1}$$

As for why substituting values of $x = 1$ and $x = 2$ works, this is because \eqref{eq1} is an identity and, thus, must be true for all values of $x$. With the original equation involving the denominators, due to continuity for all $x \neq 1,2$, the numerators must still match values at $x = 1,2$. Thus, by using $x = 2$ first, you eliminate the $A$ parameter so you only have the $B$ parameter, to get $B = 5$. Likewise, using $x = 1$ eliminates the $B$ parameter, leaving an equation in just $A$ to solve to get $-A = 3 \implies A = -3$. However, this method doesn't always work well in more complicated sets of equations (e.g., where higher powers are involved, there are considerably more variables being used so you can't isolate just one of them, etc.), which is why I present the more general method below.

Although it's sometimes more work, you can also expand and collect the terms of the same powers of $x$ together. In this case, \eqref{eq1} becomes

$$2x + 1 = (A + B)x - 2A - B \implies (2 - (A + B))x + (1 + 2A + B) = 0 \tag{2}\label{eq2}$$

Thus, you get for the coefficient of $x$ to be $0$ that

$$A + B = 2 \tag{3}\label{eq3}$$

and for the constant term to be $0$ that

$$-2A - B = 1 \tag{4}\label{eq4}$$

Adding the $2$ equations gives $-A = 3 \implies A = -3$. Thus, from \eqref{eq3}, you then get $B = 2 + 3 = 5$. This, of course, matches what was originally determined by using $x = 1$ and $x = 2$.

But how can we put $x=1, 2$ as the identity has come from eliminating the denominator containing $(x-1)(x-2)$ — user679770, Aug 08 '19 at 17:45
@user679770 I'm sorry, but my earlier response was not very accurate, so I deleted it. As the comment by InterstellarProbe indicates, at x=1 and x=2, the fractions still need to work. The way I look at it is $\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$ is true for all $x \neq 1,2$. However, by continuity, the numerators must still match values also at $x = 1,2$. This is how you can determine what $A$ and $B$ must be, as you ask about & I explicitly show in my answer above. — John Omielan, Aug 08 '19 at 18:31

score 1 · Answer 2 · answered Aug 08 '19 at 17:49

\begin{align*} \frac{2x+1}{(x-1)(x-2)} & = \frac{(2x-2) + 3}{(x-1)(x-2)} = \frac{2(x-1)}{(x-1)(x-2)} + \frac{3}{(x-1)(x-2)}\\\\ & = \frac{2}{x-2} + 3\times\frac{(x-1) - (x-2)}{(x-1)(x-2)} = \frac{2}{x-2} + \frac{3}{x-2} - \frac{3}{x-1}\\\\ & = \frac{5}{x-2} - \frac{3}{x-1} \end{align*}

score 0 · Answer 3 · answered Aug 08 '19 at 17:52

While $\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$ is only valid for $x\not\in\{1,\,2\}$ for the values of $A,\,B$ we seek, $2x+1=A(x-2)+B(x-1)$ will be valid for all $x\in\Bbb R$ because its validity at other values implies validity on $x\in\{1,\,2\}$ by continuity. You could also find $A,\,B$ from simultaneous equations obtained at other values of $x$, but using $x\in\{1,\,2\}$ is especially convenient, viz. $3=-A,\,5=B$. Or you could just equate coefficients of $x^1=x,\,x^0=1$, viz. $A+B=2,\,-2A-B=1$. That approach will serve you well for other kinds of coefficient-inferring problems.

score 0 · Answer 4 · answered Aug 08 '19 at 18:26

The expression (2x+1)/(x-1)(x-2). = A / ( x-1). + B/(x-2) is discontinuous at x= 1 and at x= 2 but identy for restall value of x.

Once we write as 2x +1 = A (x-2) + B (x-1) since left hand side is continuous thus Right-hand side will be having same behaviour .hence we can put x= 2 and x= 1 to find A and B

Computation Of Integrals

4 Answers4