Question

$\int \frac{4x+3}{\sqrt{2x^2+3x+1}}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{4x+3}{\sqrt{2x^2+3x+1}}\right)dx \\ \mathrm{Let}2x^2+3x+1=t \\ \Rightarrow \left(4x+3\right)=\frac{dt}{dx} \\ \Rightarrow \left(4x+3\right)dx=dt \\ \mathrm{Now},\int \left(\frac{4x+3}{\sqrt{2x^2+3x+1}}\right)dx \\ =\int \frac{dt}{\sqrt{t}} \\ =\int t^{-\frac{1}{2}}dt \\ =\left[\frac{t^{{\displaystyle -\frac{1}{2}}+1}}{{\displaystyle -\frac{1}{2}}+1}\right]+\mathrm{C} \\ =2\sqrt{t}+\mathrm{C} \\ =2\sqrt{2x^2+3x+1}+\mathrm{C} \end{gathered}$$