Integration by anti-differentiation

Question

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 $\heartsuit$

Answer 1

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 $2(1 - t)^{-1/2} + c$.