Mathematics

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


ANSWER

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


SOLUTION
Given,

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

put $$u=4-3x$$

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

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

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

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

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

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
Evaluate : $$\displaystyle \int \dfrac { cos 2x}{cos^2 x sin^2 x }dx  $$
  • A. $$ -cot x + tan x + C$$
  • B. $$ cot x -tan x +C $$
  • C. $$ cot x + tan x + C$$
  • D. $$ - cot x -tan x +C $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 One Word Hard
Evaluate $$\displaystyle \int \frac{e^{x}}{e^{2x}+6e^{x}+5}dx.$$
The ans is $$\displaystyle =\frac{1}{4}log\left | \frac{e^{x}+1}{e^{x}+ k} \right |+C$$
what is k

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
$$\int {{2^x}.{e^x}dx = } $$
  • A. $${{{2^x}.{e^x}} \over {1 - \ell n2}} + C$$
  • B. $${{{2^x}.{e^x}} \over { - 1 + \ell n2}} + C$$
  • C. $${{{{\left( {2e} \right)}^x}} \over {\ell n\left( {2e} \right)}} + C$$
  • D. $${{{2^x}.{e^x}} \over {1 + \ell n2}} + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Calculate the following integrals.
$$\displaystyle \int_{0}^{2e} \, \frac{dx}{0.5x \, + \, 1}.$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate:
$$ \int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer