Mathematics

# $\int_0^1 {\dfrac{{dx}}{{x + \sqrt x }}}$ equals

$2ln\ 2$

##### SOLUTION

Consider the given integral.

$I=\int_{0}^{1}{\dfrac{dx}{x+\sqrt{x}}}$

$I=\int_{0}^{1}{\dfrac{dx}{\sqrt{x}\left( 1+\sqrt{x} \right)}}$

Let $t=1+\sqrt{x}$

$\dfrac{dt}{dx}=\dfrac{1}{2\sqrt{x}}$

$2dt=\dfrac{1}{\sqrt{x}}dx$

Therefore,

$I=2\int_{1}^{2}{\dfrac{dt}{t}}$

$I=2\left[ \ln \left( t \right) \right]_{1}^{2}$

$I=2\left[ \ln \left( 2 \right)-\ln \left( 1 \right) \right]$

$I=2\ln \left( 2 \right)$

Hence, the value of integral is $2\ln \left( 2 \right)$.

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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
$\displaystyle \int { \frac { \sec { x } }{ \sqrt { \sin { \left(2 x+\alpha \right) } +\sin { \alpha } } } dx } =$
• A. $\displaystyle \sqrt { 2\left( \sin { x } +\tan { \alpha } \right) \sec { \alpha } }$
• B. $\displaystyle \sqrt { \left( \tan { x } +\tan { \alpha } \right) \sec { \alpha } }$
• C. None of these
• D. $\displaystyle \sqrt { 2\left( \tan { x } +\tan { \alpha } \right) \sec { \alpha } }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate : $\displaystyle \int \frac{x^{3}+x+1}{x^{2}-1}dx$
• A. $\displaystyle =\frac{x^{3}}{2}+log\left | x^{3}-1 \right |+\frac{1}{2}log\left |\frac{x-1}{x+1} \right |+C$
• B. $\displaystyle =\frac{3x^{2}}{2}+log\left | 3x^{2}-1 \right |+\frac{1}{2}log\left |\frac{x-1}{x+1} \right |+C$
• C. $\displaystyle =\frac{x^{2}}{5}+log\left | x^{2}+1 \right |+\frac{1}{3}log\left |\frac{x-1}{x+1} \right |+C$
• D. $\displaystyle =\frac{x^{2}}{2}+log\left | x^{2}-1 \right |+\frac{1}{2}log\left |\frac{x-1}{x+1} \right |+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate: $\displaystyle \int_{0}^{1} \dfrac {dx}{x +\sqrt {1 - x^{2}}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int\frac{\tan x}{\sec x+\cos x}dx=$
• A. $\tan^{-1}(\cos x)+c$
• B. $\tan^{-1} (\sin x)+c$
• C. $-\tan^{-1}(\sin x)+c$
• D. $-\tan^{-1}(\cos x)+c$

$\displaystyle\int\limits_{a}^{b}f(x)\ dx=b^3-a^3$, then find $f(x)$.