Mathematics

# Write a value of $\displaystyle\int { { e }^{ x }\left( \cfrac { 1 }{ x } -\cfrac { 1 }{ { x }^{ 2 } } \right) } dx$

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

$=\displaystyle\int{{e}^{x}\dfrac{1}{x}dx}-\displaystyle\int{{e}^{x}\dfrac{1}{{x}^{2}}dx}$

Let $u={e}^{x}\Rightarrow\,du={e}^{x}dx$

$dv=\dfrac{1}{{x}^{2}}dx\Rightarrow\,v=\dfrac{-1}{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$......by parts formula.

Integrating by parts, we get

$=\displaystyle\int{{e}^{x}\dfrac{1}{x}dx}+\dfrac{1}{x}{e}^{x}-\displaystyle\int{{e}^{x}\dfrac{1}{x}dx}+c$

$=\dfrac{1}{x}{e}^{x}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
If $\int_{-1}^{1}\displaystyle \frac{x}{\sqrt{1-x^{2}}} sin^{-1}(2x \sqrt{1-x^{2}}) dx = k(\sqrt{2}-1) ,$ then find the value of $k$.

• A. $2$
• B. $\sqrt { 2 } -1$
• C. $\sqrt { 2 } +1$
• D. $4$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following integral:
$\displaystyle\int^{\pi/2}_0\dfrac{dx}{(1+\cos^2x)}$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
A cubic function $\displaystyle f(x)$ vanishes at $\displaystyle x=-2$ and has a relative minima/maxima at $\displaystyle x=1$ and $\displaystyle x=1/3$ if $\displaystyle \int_{-1}^{1}f\left ( x \right )dx= \frac{14}{3}$ then $\displaystyle f(x)$ equals
• A. $\displaystyle \left ( x+2 \right )\left ( x^{2}+x-1 \right )$
• B. $\displaystyle \left ( x+3 \right )\left ( x+2 \right )\left ( x-1 \right )$
• C. None of these
• D. $\displaystyle \left ( x+2 \right )\left ( x^{2}-x+1 \right )$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Integrate $\displaystyle \int \dfrac {\log x}{x^{2}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$