Mathematics

# $\displaystyle \int x \frac{\ln (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}\, dx$ equals to

$\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$

##### SOLUTION
$I = \displaystyle \int x \frac{\ln (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}\, dx$

$\quad = \displaystyle \int \frac{2x}{2\sqrt{1 + x^{2}}}.\ln (x + \sqrt{1 + x^{2}}) dx$

Let
$I_1 = \displaystyle \int \frac{2x}{2\sqrt{1 + x^{2}}}.$

Put $1+x^2=t\implies2x dx=dt$

$I_1=\displaystyle \int \dfrac{1}{2}t^{\tfrac{-1}{2}}dt=t^{\tfrac{1}{2}}= \sqrt{1 + x^{2}}$

Using integral by parts,
$\displaystyle \int u.v \ dx=u\int v\ dx$ $\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$I = \displaystyle \ln (x + \sqrt{1 + x^{2}}) \int \frac{2x}{2\sqrt{1 + x^{2}}} dx -\int \left[ \frac{d\{\ln (x + \sqrt{1 + x^{2}})\}}{dx}. \int \frac{2x}{2\sqrt{1 + x^{2}}} dx\right ] dx+c$

$\quad = \displaystyle \ln (x + \sqrt{1 + x^{2}}). \sqrt{1 + x^{2}} -\int \left[ \frac{1}{x + \sqrt{1 + x^{2}}}. \left(1+\frac{x}{\sqrt{1+x^2}} \right).\sqrt{1 + x^{2}} \right ] dx+c$

$\quad = \displaystyle \ln (x + \sqrt{1 + x^{2}}). \sqrt{1 + x^{2}} -\int dx+c=\ln (x + \sqrt{1 + x^{2}}). \sqrt{1 + x^{2}} -x+c$

$\quad =\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Hard
Solve $\displaystyle \int \dfrac{1}{x^4+x^2+1}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle\int { \cfrac { \sqrt { x } }{ \sqrt { x } -\sqrt [ 3 ]{ x } } } dx$ is equal to
• A. $6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 2 } +\cfrac { { x }^{ 1/3 } }{ 3 } +\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$
• B. $6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 3 } +\cfrac { { x }^{ 1/3 } }{ 2 } +\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$
• C. $6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 2 } +\cfrac { { x }^{ 1/3 } }{ 3 } +{ x }^{ 1/6 }+\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$
• D. None of the above

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Integrate $\displaystyle \int _{0}^{2/3}\dfrac{dx}{4+9x^{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\displaystyle \int {\dfrac{{d({x^2} + 1)}}{{\sqrt {{x^2} + 2} }}} ,$ is
• A. $\sqrt {{x^2} + 2} + c$
• B. $x\sqrt {{x^2} + 2} + c\,\,\,\,\,\,$
• C. $4\sqrt {{x^2} + 2} + c$
• D. $2\sqrt {{x^2} + 2} + c$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$