What is the integral of 1 over t times the quantity t squared minus 4?

[karush]

{W8.7.7} nmh{1000}
$\displaystyle
I=\int \frac{1}{t\left(t^2 - 4 \right)}
\ d{t}

= \frac{1}{8}\ln\left|{1-\frac{4}{{t}^{2}}}\right|+ C$

$$I=\int \frac{1}{t\left(t^2 - 4 \right)}
\ d{t}=-\int\frac{1}{t\left(4-{t}^{2 } \right)}
\ d{t}$$
$\begin{align}\displaystyle
t& = 2\sin\left(u\right)&
dt&= 2\cos\left({u}\right)\ d{u}
\end{align}$

$$\int\frac{2\cos\left({u}\right)}
{2 \sin\left({u}\right)4\cos^2 \left({u}\right)} \ du $$

$\text{I continued but didn't seem to work}$

 

[MarkFL]

What you can do is:

$$I=\int\frac{1}{t(t^2-4)}\,dt=\frac{1}{8}\int\frac{8t^{-3}}{1-4t^{-2}}\,dt$$

Now use a $u$-subsititution...

 

[karush]

$\displaystyle
I=\frac{1}{8}\int\frac{8t^{-3}}{1-4t^{-2}}\,dt
=\frac{1}{8}\int\frac
{\frac{8}{t^3}}
{1-\frac{4}{{t}^{2}}}
$Then substitute $2\sin(u)$ for $t$?

 

[MarkFL]

$\displaystyle
I=\frac{1}{8}\int\frac{8t^{-3}}{1-4t^{-2}}\,dt
=\frac{1}{8}\int\frac
{\frac{8}{t^3}}
{1-\frac{4}{{t}^{2}}}
$Then substitute $2\sin(u)$ for $t$?

No, let:

$$u=1-4t^{-2}\,\therefore\,du=8t^{-3}\,dt$$

 

[topsquark]

$\tiny\text{Whitman 8.7.7 inverse integral} $
$$\displaystyle
I=\int \frac{1}{t\left(t^2 - 4 \right)}
\ d{t}

= \frac{1}{8}\ln\left|{1-\frac{4}{{t}^{2}}}\right|+ C$$

$$I=\int \frac{1}{t\left(t^2 - 4 \right)}
\ d{t}=-\int\frac{1}{t\left(4-{t}^{2 } \right)}
\ d{t}$$
$\begin{align}\displaystyle
t& = 2\sin\left(u\right)&
dt&= 2\cos\left({u}\right)\ d{u}
\end{align}$

$$\int\frac{2\cos\left({u}\right)}
{2 \sin\left({u}\right)4\cos^2 \left({u}\right)} \ du $$

$\text{I continued but didn't seem to work}$$\tiny\text
{from Surf the Nations math study group} \\
$

I don't know what methods you are working with but you could also do partial fractions:
[math]\frac{1}{t(t^2 - 4)} = \frac{1}{8} \left ( \frac{-2}{t} + \frac{1}{t + 2} + \frac{1}{t - 2} \right )[/math]

-Dan

 

[MarkFL]

I don't know what methods you are working with but you could also do partial fractions:
[math]\frac{1}{t(t^2 - 4)} = \frac{1}{8} \left ( \frac{-2}{t} + \frac{1}{t + 2} + \frac{1}{t - 2} \right )[/math]

-Dan

Without having seen the anti-derivative given in the first post, that's probably how I would have approached the integral. :)

 

[karush]

No, let:

$$u=1-4t^{-2}\,\therefore\,du=8t^{-3}\,dt$$

So then
$$I=\frac{1}{8}\int\frac{1}{u} \ du
= \frac{1}{8}\ln\left|{1-\frac{4}{{t}^{2 }}}\right|+C$$

I guess the most obvious is the hardest thing to see!