Mathematics

# The value of $\int { { e }^{ 2x } } \left(\dfrac{1}{x} - \dfrac{1}{2x^2}\right)$ dx is

##### SOLUTION
Let $2x=t$
$\Rightarrow 2dx=dt$
$\Rightarrow I=\displaystyle\int e^t\left(\dfrac{2}{t}-\dfrac{4}{2t^2}\right)\dfrac{dt}{2}$
$=\displaystyle\int e^t\left(\dfrac{1}{t}-\dfrac{1}{t^2}\right)dt$
$=$ of form $\displaystyle\int e^x(f(x)+f'(x))dx$
$=e^xf(x)+c$
$\Rightarrow I=e^t\left(\dfrac{1}{t}\right)+c$
$\Rightarrow I=\dfrac{e^{2x}}{2x}+c$

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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 Hard
Let $[x]$ denote the largest integer not exceeding $x$ and $\left \{x\right \} = x - [x]$. Then
$\int_{0}^{2012} \dfrac {e^{\cos(\pi \left \{x\right \})}}{e^{\cos(\pi \left \{x\right \})} + e^{-\cos(\pi \left \{x\right \})}} dx$ is equal to.
• A. $0$
• B. $2012$
• C. $2012\pi$
• D. $1006$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int { \frac { 1 }{ 1-\cos ^{ 4 }{ x } } dx } =-\frac { 1 }{ 2\tan { x } } +\frac { k }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \frac { \tan { x } }{ \sqrt { 2 } } \right) } +C$, where $k=$
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• B. $-1$
• C. $1$
• D. $\displaystyle \frac { 1 }{ 2 }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate: $\displaystyle \int e^{2x} \sin 3x \ dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate
$\displaystyle \int{\dfrac{x^{2}\sin^{-1}x}{(1-x^{2})^{3/2}}dx}$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$