I am currently trying to find the arc length of $f(x)=ln(x)$, which involves the integral $$\int \sqrt{1+\frac1{x^2}}dx$$ I managed to solve the integral correctly but I want to know if there is a simpler way as I ended up canceling two terms later on.
For reference this is what I did: $$\int \sqrt{1+\frac1{x^2}}dx$$ $$\int \frac1x\sqrt{x^2+1}dx$$ $$u_1=\frac1x$$
$$du_1=-\frac{dx}{x^2}$$
$$v_1=\int \sqrt{x^2+1}dx$$
$$v_1=\frac12(sinh^{-1}(x)+x\sqrt{x^2+1})$$
$$\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)+x\sqrt{x^2+1}}{2x}+\frac12\int\frac{sinh^{-1}(x)}{x^2}dx+\int\frac1x\sqrt{x^2+1}dx$$ $$\frac12\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)+x\sqrt{x^2+1}}{2x}+\frac12\int\frac{sinh^{-1}(x)}{x^2}dx$$ $$\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)+x\sqrt{x^2+1}}x+\int\frac{sinh^{-1}(x)}{x^2}dx$$ $$u_2=sinh^{-1}(x)$$ $$du_2=\frac{dx}{\sqrt{x^2+1}}$$ $$v_2=\int\frac{dx}{x^2}$$ $$v_2=\frac1x$$ $$\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)}x+{\sqrt{x^2+1}}-\frac{sinh^{-1}(x)}x+\int\frac{dx}{x\sqrt{x^2+1}}$$ $$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}+\int\frac{dx}{x\sqrt{x^2+1}}$$ $$t=\sqrt{1+x^2}$$ $$t^2-1=x^2$$ $$dt= \frac x{\sqrt{x^2+1}} dx$$ $$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}+\int\frac x{x^2\sqrt{x^2+1}}dx$$ $$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}+\int\frac{dt}{t^2-1}$$ $$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}-tanh^{-1}(t)+C$$ $$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}-tanh^{-1}(\sqrt{1+x^2})+C$$ Is there a simpler way to do this question instead of what I did?
Standard ways to find this integral are to substitute $x=\tan\theta$ or $x=\sinh t$.
You can also let $\frac{1}{x}=\tan\theta$, so $x=\cot\theta$ and
$\displaystyle\int\sqrt{1+\frac{1}{x^2}}\;dx=\int\sec\theta\big(-\csc^{2}\theta\big)d\theta$. $\;\;$Then letting $u=\sec\theta, dv=-\csc^{2}\theta d\theta$ gives
$\displaystyle\sec\theta\cot\theta-\int\sec\theta d\theta=\csc\theta-\ln\lvert\sec\theta+\tan\theta\rvert+C=\sqrt{x^2+1}-\ln\bigg\vert\frac{\sqrt{x^2+1}+1}{x}\bigg\vert+C$