Find the length of the parametric curve Find the length of the parametric curve defined by:
$x=t+\dfrac{1}{t}$
and
$y=\ln{t^2}$ 
on the interval
$(1 \le t \le 4)$.
 A: Hint: The derivative of your first function is $1-\frac{1}{t^2}$. The derivative of the second is $\frac{2}{t}$. Note that
the sum of the squares of the derivatives is 
$$\left(1+\frac{1}{t^2}\right)^2.$$
Now we can take the square root easily, and integrate.
Added: We have, using the familiar $(a+b)^2=a^2+2ab+b^2$, 
$$\left(1-\frac{1}{t^2}\right)^2=1-\frac{2}{t^2}+\frac{1}{t^4}.$$
Add the square of $\frac{2}{t}$, that is, $\frac{4}{t^2}$.  We get
$$1+\frac{2}{t^2}+\frac{1}{t^4},$$
which is equal to $\left(1+\frac{1}{t^2}\right)^2$.
Remark: This sort of trickery happens often in arclength problems, parametric or not. The point is that if we take two reasonable functions $x(t),y(t)$, and calculate 
$$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},$$
we typically get a function which has no elementary antiderivative. Thus many of the arclength problems in calculus books involve very specially selected functions for which there is "magic" cancellation that makes the thing inside the square root the square of something nice. Note that if we change the function a little bit, taking in our case $x(t)=t+\frac{1.2}{t}$ we end up with an integral that cannot be done in terms of elementary functions.  
A: $$S=\int_1^4\sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}dt$$
where
$$\frac{dx}{dt}=\frac{d}{dt}\bigg(t+\frac 1t\bigg)=1-\frac 1{t^2}$$
$$\frac{dy}{dt}=\frac{d}{dt}\bigg(\ln(t^2)\bigg)=\frac 2{t}$$
and
$$S=\int_1^4\sqrt{\bigg(t+\frac 1t\bigg)^2+\bigg(\frac 2{t}\bigg)^2}dt=\int_1^4{\frac{(1+t^2)}{t^2}}dt=\bigg(t-\frac 1t\bigg)\bigg|_1^4$$
and
$$S=\frac{15}4$$
A: $$\begin{align*}s&=\int_1^4\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\\&=\int_1^4\sqrt{\left(1-\frac1{t^2}\right)^2+\left(\frac2t\right)^2}dt\\&=\int_1^4\sqrt{1-\frac2{t^2}+\frac1{t^4}+\frac4{t^2}}dt\\&=\int_1^4\sqrt{\left(\frac1{t^2}\right)^2+2\left(\frac1{t^2}\right)+1}dt\\&=\int_1^4\sqrt{\left(\frac1{t^2}+1\right)^2}dt\\&=\int_1^4\left(\frac1{t^2}+1\right)dt\\&=\left[-\frac1t+t\right]_1^4\\&=-\frac14+4+1-1\\&=\frac{15}4\end{align*}$$
