Mathematics

# Integrate:$\int { \dfrac { \left( x-3 \right) { e }^{ x } }{ { \left( x-1 \right) }^{ 3 } } dx }$

##### SOLUTION
$\displaystyle \int \dfrac{(x-3)e^{x}}{(x-1)^{3}}dx$ (By ILATE)
let $x-1 = m$ or $m+1 = x$
$dx =dm$
$\Rightarrow \displaystyle \int \dfrac{(m-2)e^{m+1}}{m^{3}}dm$
Here (1) $u = (m-2)e^{m+1}, V = \dfrac{1}{m^{3}}$
$(\because \displaystyle \int uv \, dx = u \int v dx - \int \left [ \dfrac{dv}{dx} \int v \,dx \right ])$
$\Rightarrow \dfrac{-e^{m+1}(m-2)}{2m^{2}} - \displaystyle \int -\frac{e^{m+1}+e^{m+1}m}{2m^{2}}dm$
$\displaystyle -\dfrac{1}{2} \int \dfrac{-e^{m+1}+e^{m+1}m}{m^{2}}dm$
Here (2)$m = e^{m+1}+e^{m+1} m , v = \dfrac{1}{m^{2}}$
$\Rightarrow \frac{-1}{2} \left [ -\left ( \dfrac{-e^{m+1}+e^{m+1}m}{m} \right )-e^{m+1} dn \right ]$
$(-e^{m+1})$
$\Rightarrow \dfrac{-1}{2} \left ( -\dfrac{e^{m+1}+e^{m+1}m}{m} -(-e^{1+m})\right )$
$\dfrac{-e^{m+1}}{2m}$
$\Rightarrow \dfrac{-e^{m+1}(m-2)}{2m^{2}} + \dfrac{e^{m+1}}{2m}+c$
or
$\Rightarrow \boxed{ \dfrac{e^{x}(x-3)}{2(x-1)^{2}}+ \dfrac{e^{x}}{2(x-1)}+c}$

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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
$\displaystyle\int { \dfrac { \left( 1+x \right) { e }^{ x } }{ \cot { \left( x{ e }^{ x } \right) } } dx }$ is equal to
• A. $\log { \left| \cos { \left( x{ e }^{ x } \right) } \right| } +C$
• B. $\log { \left| \cot { \left( x{ e }^{ x } \right) } \right| } +C$
• C. $\log { \left| \cos { \left( x{ e }^{ -x } \right) } \right| } +C$
• D. $\log { \left| \sec { \left( x{ e }^{ x } \right) } \right| } +C$
• E. $\log { \left| \sec { \left( x{ e }^{ -x } \right) } \right| } +C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle\int_{0}^{1}\sqrt x \ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle \int_{-1}^{1}x|x|dx$ is
• A. $2$
• B. $1$
• C. None of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Prove that $\displaystyle\int^{\pi/4}_{-\pi/4}|\sin x|dx=(2-\sqrt{2})$.

$\int {\cos \sqrt x dx\,.}$