Mathematics

# $\displaystyle \int { \dfrac { 1 }{ \sqrt { 4-3x } } } dx=\_ \_ \_ \_ \_ \_ +c$

$-\frac { 2 }{ 3 } { (4-3x) }^{ \frac { 1 }{ 2 } }$

##### SOLUTION
Given,

$\displaystyle \int \dfrac{1}{\sqrt{4-3x}}dx$

put $u=4-3x$

$\displaystyle=\int \:-\dfrac{1}{3\sqrt{u}}du$

$\displaystyle=-\dfrac{1}{3}\cdot \int \:u^{-\frac{1}{2}}du$

$=-\dfrac{1}{3}\cdot \dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$

$=-\dfrac{1}{3}\cdot \dfrac{\left(4-3x\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$

$=-\dfrac{2}{3}\sqrt{4-3x}+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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