integral of dx/(3x^2+1) with u substitution
- Thread starter iamjesse
- Start date
You could use the substitution:
\(\displaystyle \L\\x=\frac{tan(u)}{\sqrt{3}};\;\ dx=\frac{1}{\sqrt{3}}sec^{2}(u)du\)
Make the substitutions:
\(\displaystyle \L\\\int\frac{1}{3(\frac{tan(u)}{\sqrt{3}})^{2}+1}(\frac{1}{\sqrt{3}}sec^{2}(u))\)
This reduces nicely to:
\(\displaystyle \L\\\int\frac{1}{\sqrt{3}}du\)
Integrating we get:
\(\displaystyle \L\\\int\frac{1}{\sqrt{3}}u\)
Resub:
\(\displaystyle \H\\\frac{1}{\sqrt{3}}tan^{-1}(\sqrt{3}x)+C\)
\(\displaystyle \L\\x=\frac{tan(u)}{\sqrt{3}};\;\ dx=\frac{1}{\sqrt{3}}sec^{2}(u)du\)
Make the substitutions:
\(\displaystyle \L\\\int\frac{1}{3(\frac{tan(u)}{\sqrt{3}})^{2}+1}(\frac{1}{\sqrt{3}}sec^{2}(u))\)
This reduces nicely to:
\(\displaystyle \L\\\int\frac{1}{\sqrt{3}}du\)
Integrating we get:
\(\displaystyle \L\\\int\frac{1}{\sqrt{3}}u\)
Resub:
\(\displaystyle \H\\\frac{1}{\sqrt{3}}tan^{-1}(\sqrt{3}x)+C\)