English

∫ 1 √ 5 X 2 − 2 X D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int\frac{dx}{\sqrt{5 x^2 - 2x}}\]
\[ = \int\frac{dx}{\sqrt{5\left( x^2 - \frac{2}{5}x \right)}}\]
\[ = \frac{1}{\sqrt{5}}\int\frac{dx}{\sqrt{x^2 - \frac{2}{5}x + \left( \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^2}}\]
\[ = \frac{1}{\sqrt{5}}\int\frac{dx}{\sqrt{\left( x - \frac{1}{5} \right)^2 - \left( \frac{1}{5} \right)^2}}\]
\[ = \frac{1}{\sqrt{5}} \text{ log }\left| x - \frac{1}{5} + \sqrt{\left( x - \frac{1}{5} \right)^2 + \left( \frac{1}{5} \right)^2} \right| + C\]
\[ = \frac{1}{\sqrt{5}} \text{ log }\left| \frac{5x - 1}{5} + \frac{\sqrt{5 x^2 - 2x}}{\sqrt{5}} \right| + C\]

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 9 | Page 93

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

`int{sqrtx(ax^2+bx+c)}dx`

 
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec         } {x }- \cot x} dx\]

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

\[\int \left( e^x + 1 \right)^2 e^x dx\]

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

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

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

\[\int\sqrt{1 + e^x} .  e^x dx\]

\[\int\frac{e^\sqrt{x} \cos \left( e^\sqrt{x} \right)}{\sqrt{x}} dx\]

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]

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

\[\int \sec^4 2x \text{ dx }\]

\[\int \sin^7 x  \text{ dx }\]

\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]

\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]

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

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

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

`int 1/(cos x - sin x)dx`

\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]

\[\int2 x^3 e^{x^2} dx\]

\[\int\cos\sqrt{x}\ dx\]

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

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

\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{  dx }\]

 


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

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

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

\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]

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

Write a value of

\[\int e^{3 \text{ log x}} x^4\text{ dx}\]

If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]


\[\int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\} dx =\]
 

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then


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

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

\[\int\frac{1}{4 x^2 + 4x + 5} dx\]

\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×