Question
$\int\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\text{dx}$
✓
Answer
$\int\Big(\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+2}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3-1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3}{\sqrt{2\text{x}+3}}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\sqrt{2\text{x}+3}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\Big[\int(2\text{x}+3)^\frac{1}{2}\text{dx}-\int(2\text{x}+3)^{-\frac{1}{2}}\text{dx}\Big]$
$=\frac{1}{2}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\Big)}-\frac{(2\text{x}+3)^{-\frac{1}{2}+1}}{2\big(-\frac{1}{2}+1\big)}+\text{C}\Bigg]$
$=\frac{1}{2}\Big[\frac{1}{3}(2\text{x}+3)^\frac{3}{2}-(2\text{x}+3)^\frac{1}{2}+\text{C}\Big]$
$=\frac{1}{6}(2\text{x}+3)^\frac{3}{2}-\frac{1}{2}(2\text{x}+3)^\frac{1}{2}+\text{C}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+2}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3-1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3}{\sqrt{2\text{x}+3}}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\sqrt{2\text{x}+3}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\Big[\int(2\text{x}+3)^\frac{1}{2}\text{dx}-\int(2\text{x}+3)^{-\frac{1}{2}}\text{dx}\Big]$
$=\frac{1}{2}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\Big)}-\frac{(2\text{x}+3)^{-\frac{1}{2}+1}}{2\big(-\frac{1}{2}+1\big)}+\text{C}\Bigg]$
$=\frac{1}{2}\Big[\frac{1}{3}(2\text{x}+3)^\frac{3}{2}-(2\text{x}+3)^\frac{1}{2}+\text{C}\Big]$
$=\frac{1}{6}(2\text{x}+3)^\frac{3}{2}-\frac{1}{2}(2\text{x}+3)^\frac{1}{2}+\text{C}$
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