Mathematics

# Evaluate the following integral:$\int { \cfrac { 4x+3 }{ \sqrt { 2{ x }^{ 2 }+3x+1 } } } dx$

##### SOLUTION
Let
$t=2{x}^{2}+3x+1\Rightarrow\,dt=\left(4x+3\right)dx$

$\displaystyle\int{\dfrac{\left(4x+3\right)dx}{\sqrt{2{x}^{2}+3x+1}}}$

$=\displaystyle\int{\dfrac{dt}{\sqrt{t}}}$

$=\displaystyle\int{{t}^{\frac{-1}{2}}dt}$

$=\dfrac{{t}^{\frac{-1}{2}+1}}{\dfrac{-1}{2}+1}+c$

$=\dfrac{{t}^{\frac{1}{2}}}{\dfrac{1}{2}}+c$

$=2\sqrt{t}+c$

$=2\sqrt{2{x}^{2}+3x+1}+c$ where $t=2{x}^{2}+3x+1$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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