Mathematics

# Evaluate $\displaystyle \int \dfrac{x-3}{(x-1)^{3}} dx$

##### SOLUTION
$\displaystyle \int{\dfrac{\left(x-3\right)dx}{{\left(x-1\right)}^{3}}}$
$=\displaystyle \int{\dfrac{\left(x-1-2\right)dx}{{\left(x-1\right)}^{3}}}$
$=\displaystyle \int{\dfrac{\left(x-1\right)dx}{{\left(x-1\right)}^{3}}}-2\int{\dfrac{dx}{{\left(x-1\right)}^{3}}}$
$=\displaystyle \int{\dfrac{dx}{{\left(x-1\right)}^{2}}}-2\int{\dfrac{dx}{{\left(x-1\right)}^{3}}}$
$=\displaystyle \int{{\left(x-1\right)}^{-2}dx}-2\int{{\left(x-1\right)}^{-3}dx}$
$=\dfrac{{\left(x-1\right)}^{-2+1}}{-2+1}-2\dfrac{{\left(x-1\right)}^{-3+1}}{-3+1}$
$=\dfrac{{\left(x-1\right)}^{-1}}{-1}-2\dfrac{{\left(x-1\right)}^{-2}}{-2}$
$=\dfrac{-1}{x-1}+\dfrac{1}{{\left(x-1\right)}^{2}}+c$

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

#### Realted Questions

Q1 Single Correct Medium
The value of $\displaystyle \int \dfrac{dx}{\sqrt {x} (x + 9)} dx$ is equal to:
• A. $\tan ^{-1} \sqrt{x} + c$
• B. $\tan ^{-1} (\dfrac{\sqrt{x}}{3}) + c$
• C. $\dfrac{2}{3} \tan ^{-1} \sqrt{x} + c$
• D. $\dfrac{2}{3} \tan ^{-1} (\dfrac{\sqrt{x}}{3}) + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate the following integral:
$\int { \cfrac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I=\frac{\cot x}{\sqrt{a+b\cot^{2}x}}dx\left ( 0< a < b \right ),$ then I equals

• A. $\displaystyle \sqrt{b-a} \sin ^{-1}\left ( \sqrt{b-a}x \right )+Const$
• B. $\displaystyle \sqrt{b-a}\sin ^{-1}\left ( \sqrt{b-a}\sin x \right )+Const$
• C. $-{\sqrt{b-a}\cos^{-1}}\left ( \displaystyle \sqrt{\frac{b-a}{b}\sin x} \right )+Const$
• D. $\displaystyle \frac{1}{\sqrt{b-a}}\sin ^{-1}\left ( \sqrt{\frac{b-a}{b}\sin x} \right )+Const$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate:
$\int { \sqrt { 9+{ x }^{ 2 } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Subjective Medium

Evaluate the  integrals :

$\int\limits_1^2 {\dfrac{{{{\left( {x + 1} \right)}^2}}}{{\sqrt x }}dx}$