English

∫ ( √ X − 1 √ X ) 2 D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]
\[ = \int\left( x + \frac{1}{x} - 2 \right)dx\]
` = ∫   x dx + ∫   dx/x - 2\ ∫    x dx`
`= x^2 / 2 + In   | x|  -2 x +C`

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.02 [Page 14]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.02 | Q 6 | Page 14

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

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

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

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

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

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

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

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

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

\[\int\frac{3 x^5}{1 + x^{12}} dx\]

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

\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]

\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]

`int"x"^"n"."log"  "x"  "dx"`

\[\int \sin^3 \sqrt{x}\ dx\]

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

 


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

\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]

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

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

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

Write a value of

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

\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 


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

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

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

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

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

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


\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]


\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

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

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

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


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


Find: `int (sin2x)/sqrt(9 - cos^4x) dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×