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How do I use partial fractions for the expression $\dfrac{t+1}{2\sqrt{t}(t-1)}$? Because I have to find the integral of it... Thank you

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Is the fraction $\frac{t+1}{2\sqrt{t}(t-1)}$? –  B. S. Jul 14 '13 at 13:41
Yeah, that's it. –  darkchampionz Jul 14 '13 at 13:42
Partial fractions are a device that is helpful with rational functions. Your function isn't. I suggest first a substitution. (You will not find constants $a,b,c$ that work.) –  André Nicolas Jul 14 '13 at 13:51
It's a rational function of $\sqrt{t}$. –  Michael Hardy Jul 14 '13 at 16:43
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4 Answers

up vote 5 down vote accepted

Your given integrand isn't a rational function (yet) to use partial fractions, we must first obtain a ratio of polynomials. We can do this by substituting $u = \sqrt t$.

$$u = \sqrt t \implies t = u^2 \implies dt = 2u\,du$$

Now, substituting the above into our original integral gives us:

$$\int \frac{t+1}{2\sqrt{t}(t-1)}\,dt = \int \frac{u^2 + 1}{2u(u^2 - 1)}\,(2u\,du) =\int \dfrac {u^2 + 1}{u^2 - 1}\,du$$

Now, polynomial division, followed by "partial fractions" gives us: $$\int \left(1 + \dfrac {1}{u^2 - 1}\right)\,du = \int \left(1 + \dfrac {1}{(u-1)(u+1)}\right)\,du = \int \left(1 + \dfrac{A}{u - 1} + \dfrac B{u + 1} \right) \,du$$

Now we solve for $A, B$:

$A(u+1) + B(u - 1) = 2 \iff Au + A + Bu - B = 2 \iff (A + B)u + (A - B) = 2$

$A + B = 0$
$A - B = 2$

Adding the equations gives us $2A = 2 \iff A = 1 \implies B = -1$ and we'll have a result of the form $$u + A\ln|u-1| + B\ln |u+1| + \text{Constant}$$ $$ = \sqrt t +\ln|\sqrt t - 1| - \ln |\sqrt t + 1| + \text{Constant} $$ $$ = \sqrt t+ \ln\left|\dfrac{\sqrt t - 1}{\sqrt t+1}\right| + C$$

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Needs a TU!! +1 –  Amzoti Jul 15 '13 at 1:30
@Amzoti: Needs additional TU too. –  B. S. Jul 15 '13 at 15:45
surely $A=1,B=-1$? –  Tim Jul 20 '13 at 23:40
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We can easily do the fraction as you said, but I suggest to set $\sqrt{t}=u$ first and then think about the converted integral. Indeed, you will get the following integral instead after changing the variables:

$$\int \frac{t+1}{2\sqrt{t}(t-1)}dt\longrightarrow \int\frac{u^2+1}{2u(u^2-1)}\times 2u\,du$$

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Hint: Try writing $t-1=(\sqrt{t}-1)(\sqrt{t}+1)$ and $t+1=(t-1)+2$, and using the substitution $w=\sqrt{t}$.

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Let $x=\sqrt{t}$ so $dt =2xdx$ and

$$\int\frac{t+1}{2\sqrt{t}(t-1)}dt=\int\frac{x^2+1}{(x^2-1)}dx=\int dx+\int\frac{2}{x^2-1}=\int dx+\int \frac{dx}{x-1}-\int \frac{dx}{x+1}\\ =x+\ln|x-1|-\ln|x+1|+C=\sqrt{t}+\ln|\sqrt{t}-1|-\ln|\sqrt{t}+1|+C$$

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