Question
Evaluate the following integrals:
$\int\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\text{dx}$
$\int\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\text{dx}$
✓
Answer
$\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$\text{Let }\sqrt{2\text{x}^2+3\text{x}+1}=\text{t}$
$\Rightarrow(4\text{x}+3)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(4\text{x}+3)\text{dx}=\text{dt}$
$\text{Now,}\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{t}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{2\text{x}^2+3\text{x}+1}+\text{C}$
$\text{Let }\sqrt{2\text{x}^2+3\text{x}+1}=\text{t}$
$\Rightarrow(4\text{x}+3)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(4\text{x}+3)\text{dx}=\text{dt}$
$\text{Now,}\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{t}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{2\text{x}^2+3\text{x}+1}+\text{C}$
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