Partial fractions for $\frac{t+1}{2\sqrt{t}(t-1)}$ 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
 A: 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 $$u + A\ln|u-1| + B\ln |u+1| + \text{Constant}$$ $$ = \sqrt t +\ln|\sqrt t - 1| -  \ln |\sqrt t + 1| + \text{Constant} $$
$$ = \sqrt t+ \ln\left|\dfrac{\sqrt t - 1}{\sqrt t+1}\right| + C$$
A: 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:
$$\int \frac{t+1}{2\sqrt{t}(t-1)}dt\longrightarrow \int\frac{u^2+1}{2u(u^2-1)}\times 2u\,du$$
A: Hint: Try writing $t-1=(\sqrt{t}-1)(\sqrt{t}+1)$ and $t+1=(t-1)+2$, and using the substitution $w=\sqrt{t}$.
A: Let $x=\sqrt{t}$ so $dt =2xdx$ and
$$\int\frac{t+1}{2\sqrt{t}(t-1)}dt=\int\frac{x^2+1}{(x^2-1)}dx=\int dx+\int\frac{2}{x^2-1}=\int dx+\int \frac{dx}{x-1}-\int \frac{dx}{x+1}\\
=x+\ln|x-1|-\ln|x+1|+C=\sqrt{t}+\ln|\sqrt{t}-1|-\ln|\sqrt{t}+1|+C$$
