Mathematics

$\displaystyle \int \dfrac{dx}{\sqrt{x+1}- \sqrt{x}}$ equals

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

SOLUTION
Given,

$\int \dfrac{1}{\sqrt{x+1}-\sqrt{x}}dx$

multiplying by conjugate, we get,

$=\int \dfrac{1}{\frac{1}{\sqrt{x+1}+\sqrt{x}}}dx$

$=\int \sqrt{x+1}+\sqrt{x}dx$

$=\int \sqrt{x+1}dx+\int \sqrt{x}dx$

$=\dfrac{2}{3}\left(x+1\right)^{\frac{3}{2}}+\dfrac{2}{3}x^{\frac{3}{2}}$

$=\dfrac{2}{3}[\left(x+1\right)^{\frac{3}{2}}+x^{\frac{3}{2}}+c]$

Its FREE, you're just one step away

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

Realted Questions

Q1 Single Correct Hard
$\displaystyle \int \displaystyle \frac{3x + 2}{(x + 1)(x + 3)} dx =$
• A. $\displaystyle - \frac{7}{2}\, ln | x +1 | + \frac{1}{2}\, ln | x + 3| +c$
• B. $\displaystyle \frac{1}{2}\, ln | x +1 | - \frac{7}{2}\, ln | x + 3| +c$
• C. $\displaystyle \frac{7}{2}\, ln | x +1 | - \frac{1}{2}\, ln | x + 3| +c$
• D. $\displaystyle - \frac{1}{2}\, ln | x +1 | + \frac{7}{2}\, ln | x + 3| +c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int \frac{x}{\sqrt{\left ( 9+8x-x^{2} \right )}}dx.$
• A. $\displaystyle \sqrt{9+8x-x^{2}}-\sin ^{-1}\frac{x-4}{5}$
• B. $\displaystyle -\sqrt{9+8x+x^{2}}+\sin ^{-1}\frac{x-4}{5}$
• C. $\displaystyle -\sqrt{9-8x+x^{2}}+\sin ^{-1}\frac{x-4}{5}$
• D. $\displaystyle -\sqrt{9+8x-x^{2}}+\sin ^{-1}\frac{x-4}{5}$

1 Verified Answer | Published on 17th 09, 2020

Q3 One Word Medium
Evaluate$\displaystyle \int_{0}^{\sin ^{2}t}\sin ^{-1}\sqrt{x}dx+\int_{0}^{\cos ^{2}t}\cos ^{-1}\sqrt{x}dx = k$, then $tan(k) =?$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int \frac{co\sec^{2}x-203}{\left ( \cos x \right )^{203}}dx$
• A. $\displaystyle\frac{\cos x}{\left ( co\sec x \right )^{203}}$
• B. $\displaystyle\frac{-\tan x}{\left ( co\sec x \right )^{203}}$
• C. $\displaystyle\frac{\cot x}{\left ( \sin x \right )^{203}}$
• D. $\displaystyle \frac{-\cot x}{\left ( \cos x \right )^{203}}$

Let $g(x)=\dfrac{1}{2}\left[f(x)-f(-x)\right]$ for $-3\le x\le 3$ and $f(x)=2x^2-2x+1$ then find $\displaystyle\int_{-3}^{3}g(x) dx$.