Solve $\int \frac{5}{4x^2+3}dx$

Question

I've tried few way to resolve $\int \frac{5}{4x^2+3}dx$ but I think there's somthing I'm missing. As a first step I've took the constant out: $5\int \frac{1}{4x^2+3}dx$. Next I've thought it would be ok do somthing like $5\int \frac{1}{(2x)^2+3}dx => \frac{5}{2} \int \frac{2}{4x^2+3}$ and then $\frac{5}{2} \int \frac{2}{4x^2+3} = arctan(2x)+c$ but as I understand on internet it's pretty wrong and the solution is not crrect. I also tried to do a substituition by doing $t=4x^2+3=> x=\frac{\sqrt{t}}{2}-3$ but I don't think it's correct. Then I tried to solve it by using Symolab, usualy it do human compressible steps, and as a second step it do a substituition $x=\frac{\sqrt{3}}{2}u$. I can't really understand this.

Is this a correct or logic step? why it do this substition by put $x=\frac{\sqrt{3}}{2}u$? what's the best way to solve this?

Answer 1

If you do $x=\frac{\sqrt3}2u$ and $\mathrm dx=\frac{\sqrt3}2\,\mathrm du$, then$$\int\frac5{4x^2+3}\,\mathrm dx$$becomes$$\frac{5\sqrt3}2\int\frac1{3u^2+3}\,\mathrm du=\frac{5\sqrt3}6\int\frac1{u^2+1}\,\mathrm du.$$And now you use the fact that$$\arctan'u=\frac1{u^2+1}.$$

Answer 2

We have $$\int \frac{5}{4x^2+3}dx$$ $$=\frac{5}{3}\int\frac{1}{1+\frac{4x^2}{3}}dx=\frac{5}{3}\int\frac{1}{1+(\frac{2x}{\sqrt{3}})^2}dx$$

Now use the substitution $\frac{2x}{\sqrt{3}}=\tan u.$

Answer 3

$$\int \frac{5}{4x^2+3}dx=\frac{5}{3}\int \frac{1}{\frac{4x^2}{3}+1}dx=\frac{5\sqrt3}{6}\int \frac{1}{u^2+1}du=\frac{5\sqrt3}{6}\arctan u+c$$

$$u=\frac{2x}{\sqrt3}, du=\frac{2}{\sqrt3}dx$$

Answer 4

Answer :

$\int\limits_{}^{} \frac{5}{4x^2 +3} = \int\limits_{}^{}\frac{5}{(2x)^2 +3} =\frac{5}{3}\int\limits_{}^{}\frac{1}{(\frac{2x}{\sqrt{3} })^2 +1}=\frac{5×\sqrt{3 }}{3×2}\int\limits_{}^{} \frac{\frac{2}{\sqrt{3}}}{(\frac{2x}{\sqrt{3} }) ^2+1 }=\frac{5}{2\sqrt{3}}[arct(\frac{2x}{\sqrt{3}})+c ] $