Mathematics

# $\displaystyle \int \displaystyle \frac{xdx}{\sqrt{1+x^2+\sqrt{(1+x^2)^3}}}$ is equal to

$2\sqrt{1+\sqrt{1+x^2}+c}$

##### SOLUTION
Let $\displaystyle I=\int \frac { xdx }{ \sqrt { 1+x^{ 2 }+\sqrt { (1+x^{ 2 })^{ 3 } } } }$
Put $1+{ x }^{ 2 }=t\Rightarrow 2xdx=dt$

$\displaystyle I=\frac { 1 }{ 2 } \int { \frac { dt }{ \sqrt { t+{ t }^{ 3 } } } }$
Put $\displaystyle u=\sqrt { t } \Rightarrow du=\frac { 1 }{ 2\sqrt { t } } dt$

$\displaystyle I=\int { \frac { 1 }{ \sqrt { u+1 } } du } =2\sqrt { u+1 } \\ =2\sqrt { \sqrt { t } +1 } =2\sqrt { \sqrt { 1+{ x }^{ 2 } } +1 }$

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

#### Realted Questions

Q1 Single Correct Medium
If $0 < \alpha < 1$ and $\displaystyle I=\int _{-1}^{1} \frac{dx}{\sqrt{1-2\alpha x+\alpha^{2}}}$ then $I$ equals
• A. $1/\alpha$
• B. $2/\alpha$
• C. $3/\alpha$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(x^3+x \cos x +\tan^5 x+1)dx$ is equal to
• A. $0$
• B. $2$
• C. None of these
• D. $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate: $\displaystyle \int {\sqrt {{x^2} + 3x}}\,dx$
• A. $\left(\dfrac{x}{2}+\dfrac{3}{4}\right)\sqrt{x^2+3x}+\dfrac{9}{8}\cosh^{-1}{\left(\dfrac{2x}{3}+1\right)}+c$
• B. $\left(\dfrac{x}{2}-\dfrac{3}{4}\right)\sqrt{x^2+3x}-\dfrac{9}{8}\cosh^{-1}{\left(\dfrac{2x}{3}-1\right)}+c$
• C. $\left(\dfrac{x}{2}-\dfrac{3}{4}\right)\sqrt{x^2-3x}+\dfrac{9}{8}\cosh^{-1}{\left(\dfrac{2x}{3}-1\right)}+c$
• D. $\left(\dfrac{x}{2}+\dfrac{3}{4}\right)\sqrt{x^2+3x}-\dfrac{9}{8}\cosh^{-1}{\left(\dfrac{2x}{3}+1\right)}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\int (\frac{x^{6} -1}{x^{2} + 1}) dx$

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 $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$