Mathematics

# $\int\dfrac{1}{\sqrt{x}+\sqrt{x+1}}dx$ is equal to

$\dfrac{2}{3}[(x+1)^{3/2}-x^{3/2}]+C$

##### SOLUTION
$\begin{array}{l} I=\int { \dfrac { 1 }{ { \sqrt { x } +\sqrt { x+1 } } } dx } \\I= \int { \dfrac { 1 }{ { \sqrt { x } +\sqrt { x+1 } } } \times \dfrac { { \sqrt { x } -\sqrt { x+1 } } }{ { \sqrt { x } -\sqrt { x+1 } } } dx } \\I= \int { \dfrac { { \sqrt { x } -\sqrt { x+1 } } }{ { x-x-1 } } dx } \\ I=-1\left[ { \int { \sqrt { x } dx } I=-\int { \sqrt { x+1 } dx } } \right] \\ I=-\left[ { \dfrac { { { x^{ 3/2 } } } }{ { \dfrac { 3 }{ 2 } } } -\dfrac { { { { \left( { x+1 } \right) }^{ 3/2 } } } }{ { \dfrac { 3 }{ 2 } } } } \right] +c \\ I=\dfrac { 2 }{ 3 } \left[ { { { \left( { x+1 } \right) }^{ 3/2 } }-{ x^{ 3/2 } } } \right] +c \end{array}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Hard
The smallest interval in which value of $\int_{0}^{1} \dfrac{xdx}{x^{3}+3}$ lies
• A. $[0,1]$
• B. $[0,\dfrac{1}{2}]$
• C. $[0,\dfrac{1}{8}]$
• D. $[0,\dfrac{1}{4}]$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Find
$\int x ^ { 2 } + 2 x d x$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\displaystyle \int _2^3 (x^2+2x+5)\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate the integral
$\displaystyle \int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$
• A. $\displaystyle \frac{a^{2}}{4}$
• B. $\pi {a}^{2}$
• C. $\displaystyle \frac{\pi a^{2}}{2}$
• D. $\displaystyle \frac{\pi a^{2}}{4}$

$(x^2+1) \log x$