integrating using u-substitution

[mathishard124]

the problem reads as follows:

find (integral symbol) t/[(t^2)-4]^2 dt using u-substitution.

I made it so (t^2)-4 = u, and then found du/dt = 2t, and dt = du/2t

then, I substituted du/2t for dt, and thus got the following


(integral symbol) t/u^2 * du/2t, which simplified to 1/(2*u^2) * du.

finally, I found my answer to be 1/2*[(t^2)-4]^2. However, when I checked this on my calculator, it gave me a different answer. What did I do wrong?

 

[skeeter]

\(\displaystyle \L \int \frac{t}{(t^2 - 4)^2} dt\)

\(\displaystyle \L u = t^2 - 4\)

\(\displaystyle \L du = 2t dt\)

\(\displaystyle \L \frac{1}{2}\int \frac{2t}{(t^2 - 4)^2} dt\)

\(\displaystyle \L \frac{1}{2}\int \frac{1}{u^2} du = -\frac{1}{2u} + C\)

\(\displaystyle \L -\frac{1}{2(t^2 - 4)} + C\)

 

[pka]

My. you made a mess of it.

\(\displaystyle \L
\begin{array}{l}
\frac{{tdt}}{{\left( {t^2 - 4} \right)^2 }}\quad ,\quad u = t^2 - 4\quad ,\quad du = 2tdt \\
\frac{{tdt}}{{\left( {t^2 - 4} \right)^2 }} = \frac{{\left( {1/2} \right)du}}{{u^2 }} = \frac{1}{2}u^{ - 2} du \\
\end{array}\)