Mathematics

# If $\displaystyle f\left( x \right)=\int { \frac { { x }^{ 2 }dx }{ \left( 1+{ x }^{ 2 } \right) \left( 1+\sqrt { 1+{ x }^{ 2 } } \right) } }$ and $f\left( 0 \right) =0$, then the value of $f(1)$ is

$\displaystyle \log { \left( 1+\sqrt { 2 } \right) } -\frac { \pi }{ 4 }$

##### SOLUTION
$\displaystyle f\left( x \right) =\int { \frac { { x }^{ 2 }dx }{ \left( 1+{ x }^{ 2 } \right) \left( 1+\sqrt { 1+{ x }^{ 2 } } \right) } }$
Substitute $x=\tan { \theta } \Rightarrow dx=\sec ^{ 2 }{ \theta } d\theta =\left( 1+{ x }^{ 2 } \right) d\theta$
$\displaystyle \therefore f\left( x \right) =\int { \frac { { x }^{ 2 }dx }{ \left( 1+{ x }^{ 2 } \right) \left( 1+\sqrt { 1+{ x }^{ 2 } } \right) } } =\int { \frac { \tan ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } d\theta }{ \sec ^{ 2 }{ \theta } \left( 1+\sec { \theta } \right) } }$
$\displaystyle =\int { \frac { \tan ^{ 2 }{ \theta } d\theta }{ 1+\sec { \theta } } } =\int { \frac { \sin ^{ 2 }{ \theta } d\theta }{ \cos { \theta } \left( 1+\cos { \theta } \right) } } =\int { \frac { 1-\cos ^{ 2 }{ \theta } d\theta }{ \cos { \theta } \left( 1+\cos { \theta } \right) } }$
$\displaystyle =\int { \frac { \left( 1-\cos { \theta } \right) }{ \cos { \theta } } d\theta } =\int { \sec { \theta } d\theta } -\int { d\theta } \\ =\log { \left( x+\sqrt { 1+{ x }^{ 2 } } \right) } -\tan ^{ -1 }{ x } +c\\ f\left( 0 \right) =\log { \left( 0+\sqrt { 1+0 } \right) } -\tan ^{ -1 }{ 0 } +c\\ \Rightarrow 0=\log { 1 } -0+c\Rightarrow c=0$
$\displaystyle \therefore f\left( 1 \right) =\log { \left( 1+\sqrt { 1+{ 1 }^{ 2 } } \right) } -\tan ^{ -1 }{ 1 } =\log { \left( 1+\sqrt { 2 } \right) } -\frac { \pi }{ 4 }$

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

#### Realted Questions

Q1 Subjective Hard
Using the method of integration find area bounded by $|x|+|y|=1$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle\int _{ 0 }^{ { x }/{ 4 } }{ \dfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } dx }$ is
• A. $1+\sqrt{2}$
• B. $-11+\sqrt{2}$
• C. $-\sqrt{2}$
• D. None of these

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Let $\displaystyle \frac{d}{dx}F(x)=\frac{e^{{s}{m}{x}}}{x}$ , $x>0$. lf $\displaystyle \int_{1}^{4}\frac{3}{x}e^{{s}{m}{x}^{3}}dx=F(k)-F(1)$, then one of the possible values of ${k}$ is
• A. $16$
• B. $62$
• C. $15$
• D. $64$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int\frac{2x^{2}-5x+1}{x^{2}(x^{2}-1)}dx=$
• A. $\displaystyle \frac{1}{x}+\displaystyle log\left| \frac { x^{ 5 } }{ (x^{ 2 }-1)(x+1)^{ 2 } } \right| +$
• B. $\displaystyle \frac{1}{x}+\displaystyle \log\left| \frac { x^{ 5 } }{ (x^{ 2 }-1)(x+1)^{ 4 } } \right| +c$
• C. $\displaystyle \frac{1}{x}+\displaystyle \log\left| \frac { x^{ 5 } }{ (x^{ 2 }-1)(x+1) } \right| +c$
• D. $\displaystyle \frac{1}{x}+\displaystyle \log\left| \frac { x^{ 5 } }{ (x^{ 2 }-1)(x+1)^{ 3 } } \right| +c$

If $\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx =f(x) + c$ then $f(x) =$