Integration of Arc Length I am asked to calculate the length of $\mathbf{x}(t)=(7,t,t^2)$ on the interval $[1,3]$.
Here is my work:
$$
  \mathbf{x}'(t)=(0,1,2t)
$$$$
  \int_1^3\|\mathbf{x}'(t)\|dt=\int_1^3\sqrt{4t^2+1}dt
$$
How do I calculate $$\int_1^3\sqrt{4t^2+1}dt$$
 A: Notice, let $$I=\int_{1}^{3}\sqrt{4t^2+1}\ dt=2\int_{1}^{3} \sqrt{t^2+\frac{1}{4}}\ dt$$
using integration by parts 
one, we can find $$\color{blue}{\int \sqrt{x^2+a^2}\ dx=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)+C}$$
Hence, applying proper limits, we get $$2\int_{1}^{3} \sqrt{t^2+\left(\frac{1}{2}\right)^2}\ dt=\frac{2}{2}\left[t\sqrt{t^2+\left(\frac{1}{2}\right)^2}+\frac{1}{4}\ln\left|t+\sqrt{t^2+\left(\frac{1}{2}\right)^2}\right|\right]_{1}^{3}$$
$$=3\sqrt{\frac{37}{4}}+\frac{1}{4}\ln\left|3+\sqrt{\frac{37}{4}}\right|-\sqrt{\frac{5}{4}}-\frac{1}{4}\ln\left|1+\sqrt{\frac{5}{4}}\right|$$
$$=\color{blue}{\frac{3\sqrt{37}-\sqrt{5}}{2}+\frac{1}{4}\ln\left(\frac{6+\sqrt{37}}{2+\sqrt{5}}\right)}$$

Proof:
  let $$I=\int \sqrt{x^2+a^2}\ dx=\int \underbrace{\sqrt{x^2+a^2}}_{I}\cdot \underbrace{1}_{II}\ dx$$ using integration by parts $$I=x\sqrt{x^2+a^2}-\int \frac{2x}{2\sqrt{x^2+a^2}}x\ dx$$
  $$I=x\sqrt{x^2+a^2}-\int \frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\ dx$$
  $$I=x\sqrt{x^2+a^2}-\int \sqrt{x^2+a^2}\ dx+a^2\int \frac{1}{\sqrt{x^2+a^2}}\ dx$$
  $$I=x\sqrt{x^2+a^2}-I+a^2\int \frac{1}{\sqrt{x^2+a^2}}\ dx$$$$\implies I=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\int \frac{1}{\sqrt{x^2+a^2}}\ dx\right)\tag 1$$
  Now, let $x=a\tan\theta\implies dx=a\sec^2\theta\ d\theta$ hence, 
  $$\color{red}{\int \frac{1}{\sqrt{x^2+a^2}}\ dx}=\int\frac{a\sec^2\theta\ d\theta}{\sqrt{a^2\tan^2\theta+a^2}}=\int\frac{a\sec^2\theta\ d\theta}{a\sec\theta}=\int \sec\theta\ d\theta$$$$=\ln\left|\sec\theta+\tan\theta\right|+c=\ln\left|\tan\theta+\sqrt{\tan^2\theta+1}\right|+c=\ln\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2+1}\right|+c$$ $$=\ln\left|x+\sqrt{x^2+a^2}\right|+\ln|a|+c=\color{red}{\ln\left|x+\sqrt{x^2+a^2}\right|+c_1}$$
  now, setting the value of integral: $\int\frac{1}{\sqrt{x^2+a^2}}\ dx$ in (1), we get $$I=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)+C$$
  hence, we get
  $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\sqrt{x^2+a^2}\ dx=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)+C}}$$

