English

∫ X 5 + X − 2 + 2 X 2 D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int\left( \frac{x^5 + x^{- 2} + 2}{x^2} \right)dx\]
\[ = \int \left( \frac{x^5}{x^2} + \frac{x^{- 2}}{x^2} + \frac{2}{x^2} \right)dx\]
\[ = \int\left( x^3 + x^{- 4} + 2 x^{- 2} \right)dx\]
\[ = \frac{x^{3 + 1}}{3 + 1} + \frac{x^{- 4 + 1}}{- 4 + 1} + 2\frac{x^{- 2 + 1}}{- 2 + 1} + C\]
\[ = \frac{x^4}{4} - \frac{1}{3 x^3} - \frac{2}{x} + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.02 [Page 15]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.02 | Q 17 | Page 15

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

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

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

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

\[\int \sin^2\text{ b x dx}\]

` ∫  {sin 2x} /{a cos^2  x  + b sin^2  x }  ` dx 


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

\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]

 ` ∫   1 /{x^{1/3} ( x^{1/3} -1)}   ` dx


\[\int \sin^4 x \cos^3 x \text{ dx }\]

\[\int\frac{1}{x^2 - 10x + 34} dx\]

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

\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]

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

\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]

\[\int x e^{2x} \text{ dx }\]

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

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

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

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

\[\int\frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \text{ dx }\]

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

\[\int\frac{1}{7 + 5 \cos x} dx =\]

\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

\[\int \sin^4 2x\ dx\]

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

\[\int \cot^5 x\ dx\]

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

\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]


\[\int \sec^4 x\ dx\]


\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]

\[\int\sqrt{a^2 - x^2}\text{  dx }\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×