Mathematics

# $\int\dfrac{dx}{\sqrt{1-x^2}-1}=$

$\dfrac{1+\sqrt{1-x^2}}{x}-2tan^{-1}\sqrt{\dfrac{1-x}{1+x}}+C$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle I = \int \frac {dx}{(2 \sin x + \sec x)^4}$, then I equals
• A. $\displaystyle \frac {1}{5 \tan^5 x} + \frac {1}{3 \tan^6 x} - \frac {I}{(2 \sin x + \sec x)^3} + C$
• B. $\displaystyle \frac {-1}{3(2 \sin x + \sec)^3} + \tan^{-1} (3\sqrt {\tan x}) + C$
• C. $\displaystyle \frac {-1}{3(2 \sin x + \sec x)^3} - \tan^{-1} (3\sqrt {\tan x}) + C$
• D. $\displaystyle -\frac {1}{5 \tan^5 x} + \frac {1}{3 \tan^6 x} - \frac {2}{7 \tan^7 x} + C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve:
$\int \ \sqrt{x^{2}+4x+1} \ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Let $f(x)$ be a positive function. Let
$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$,
$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$,
where $2k - 1 > 0$, then $\dfrac {I_{1}}{I_{2}}$ is
• A. $2$
• B. $k$
• C. $1$
• D. $\dfrac {1}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
$\displaystyle \int \cos^{-1} \left( \frac { 1- \tan^2 x}{1+ \tan^2 x } \right) dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int\frac{d{x}}{4\sin^2{x}+4\sin{x}\cos x+5\cos^{2}x}=$
• A. $\tan^{-1} (2\tan x+1)+c$
• B. $\tan^{-1}(\displaystyle \tan x+\frac{1}{2})+c$
• C. $\displaystyle \frac{1}{4}\tan (2\tan x+1)+c$
• D. $\displaystyle \frac{1}{8}\tan^{-1} (\tan x+\displaystyle \frac{1}{2})+c$