Integration by anti-differentiation I have a question in my assignment, and I highly suspect there is a typo somewhere.

Evaluate the following integral:
  $\int \frac{1}{(1-t)^{\frac{3}{2}}} dt = \frac{t}{\sqrt{1-t^2}} + c$

What I did was to perform differentiation and got the following:
$$
\begin{align*}
\frac{d}{dt}\frac{t}{\sqrt{1-t^2}} &= \frac{(0.5)(1-t^2)^{-\frac 1 2}(-2t)(t) - \sqrt{1-t^2}}
{1-t^2} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\spadesuit
\\&=\frac{-t(1-t^2)^{-\frac 1 2} - \sqrt{1-t^2}}{1-t^2}
\\&=-\sqrt{1-t^2}
\end{align*}
$$
At the same time, by integrating formally,
$$
\begin{align*}
\int \frac{1}{(1-t)^{\frac{3}{2}}} dt
&= 2\int \frac{1}{2(1-t)^{\frac{3}{2}}} dt
\\&=(1-t)^{-\frac 1 2} + c\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\heartsuit
\end{align*}
$$
Because
$$
\begin{align*}
\frac{d}{dx}(1-t)^{-\frac 1 2}
&= (-0.5)(1-t)^{-\frac 3 2}(-1)
\\&= \frac{1}{2(1-t)^{\frac 3 2}}
\end{align*}
$$
Could somebody please check my working?
UPDATE: Solved. I will preserve this version of the question to highlight all the common mistakes that careless people like me make - which is to utilize the quotient rule wrongly at $\spadesuit$ and forgetting to include the factor of $2$ at $\heartsuit$

You made an error in differentiating. It should be
\begin{align*}
\frac{d}{dt}\frac{t}{\sqrt{1-t^2}} &= \frac{\sqrt{1-t^2} - (0.5)(1-t^2)^{-\frac 1 2}(-2t)(t)}
{1-t^2}
\\&=\frac{\sqrt{1-t^2} + t^2(1-t^2)^{-\frac 1 2}}{1-t^2}
\\&=\frac{1 - t^2 + t^2}{(1 - t^2)^{3/2}}
\\&=\frac{1}{(1 - t^2)^{3/2}}.
\end{align*}
So the integral on the left is wrong as well, the $t$ should actually be $t^2$.
Also, you missed the factor of two in your answer for the integral. It should be $2(1 - t)^{-1/2} + c$.
