Mathematics

# Evaluate $\int \dfrac{{dx}}{{\sqrt {{{\left( {2 - x} \right)}^2} - 1} }}$

##### SOLUTION

Consider the given integral.

$I=\int{\dfrac{dx}{\sqrt{{{\left( 2-x \right)}^{2}}-1}}}$

Let $t=2-x$

$dt=-dx$

$dx=-dt$

Therefore,

$I=-\int{\dfrac{dt}{\sqrt{{{t}^{2}}-1}}}$

We know that,

$\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}}=\dfrac{1}{2a}\log \left| \dfrac{x-a}{x+a} \right|+c$

Therefore,

$I=-\dfrac{1}{2}\log \left| \dfrac{t-1}{t+1} \right|+c$

On putting the value of t, we get

$I=-\dfrac{1}{2}\log \left| \dfrac{2-x-1}{2-x+1} \right|+c$

$I=-\dfrac{1}{2}\log \left| \dfrac{1-x}{3-x} \right|+c$

$I=\dfrac{1}{2}\log \left| \dfrac{3-x}{1-x} \right|+c$

Hence, the value of integral is $\dfrac{1}{2}\log \left| \dfrac{3-x}{1-x} \right|+c$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Multiple Correct Medium
If $\int \frac{x\, cos\, \alpha+1 }{(x^2+2x\, cos\, \alpha+1)^{3/2}}$ $dx= \frac{x}{\sqrt{f(x) + g(x)cos\, \alpha }}+c$ then (more than one option is correct)
• A. g(2) = 2
• B. f(1) = 2
• C. f(2) = 5
• D. g(1) = 2

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int e^x\displaystyle\frac{(x-1)}{x^2}dx$ is equal to
• A. $\displaystyle\frac{e^x}{x^2}+c$
• B. $\displaystyle\frac{-e^x}{x^2}+c$
• C. $\displaystyle\frac{-e^x}{x}+c$
• D. $\displaystyle\frac{e^x}{x}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate
$\int { \dfrac { x+1 }{ { x }^{ 2 }+3x+12 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate:
$\int { \cfrac { (x+1){ \left( x+\log { x } \right) }^{ 2 } }{ x } } dx$

Evaluate $\displaystyle \int \dfrac{1}{\sqrt{9-25x^2}}dx$.