English

∫ 1 √ 5 − 4 X − 2 X 2 D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
Sum
Advertisements

Solution

\[\int\frac{dx}{\sqrt{5 - 4x - 2 x^2}}\]
\[ = \int\frac{dx}{\sqrt{2\left[ \frac{5}{2} - 2x - x^2 \right]}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - 2x - x^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x^2 + 2x \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x^2 + 2x + 1 - 1 \right)}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{5}{2} - \left( x + 1 \right)^2 + 1}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\frac{7}{2} - \left( x + 1 \right)^2}}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{7}}{\sqrt{2}} \right)^2 - \left( x + 1 \right)^2}}\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left( \frac{\left( x + 1 \right)\sqrt{2}}{\sqrt{7}} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \sin^{- 1} \left( \sqrt{\frac{2}{7}}\left( x + 1 \right) \right) + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.17 [Page 93]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.17 | Q 3 | Page 93

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\frac{\tan x}{\sec x + \tan x} dx\]

\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]

\[\int\sin x\sqrt{1 + \cos 2x} dx\]

\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]

\[\int\frac{x^2 + x + 5}{3x + 2} dx\]

\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]

\[\int \tan^{3/2} x \sec^2 \text{x dx}\]

\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]

\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]

` ∫  {1}/{a^2 x^2- b^2}dx`

\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]

 


\[\int\frac{1}{2 x^2 - x - 1} dx\]

\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]

\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]

\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]

\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]


\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]

\[\int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ dx }\]

\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]

\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]

\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]

\[\int\sqrt{2x - x^2} \text{ dx}\]

\[\int\left( x + 2 \right) \sqrt{x^2 + x + 1} \text{  dx }\]

\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]

\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]

\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]

\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]

\[\int\frac{x^3 - 1}{x^3 + x} dx\]

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]

\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]

\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]

\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]

\[\int\frac{\cos x}{\frac{1}{4} - \cos^2 x} \text{ dx }\]

\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]

\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]

\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]

\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

\[\int\frac{x}{x^3 - 1} \text{ dx}\]

\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×