English

∫ √ X ( X 3 − 2 X ) D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int\sqrt{x} \left( x^3 - \frac{2}{x} \right)dx\]
\[ = \int\left( x^\frac{7}{2} - \frac{2}{\sqrt{x}} \right)dx\]
\[ = \int\left( x^\frac{7}{2} - 2 x^{- \frac{1}{2}} \right) dx\]
`= x^(7/2+1) / (7/2+1)  - 2 (x^(-1/2+1))/(-1/2+1 )+ C`
\[ = \frac{2}{9} x^\frac{9}{2} - 4 x^\frac{1}{2} + C\]
\[ = \frac{2}{9} x^\frac{9}{2} - 4\sqrt{x} + C\]

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 10 | Page 14

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\frac{x^6 + 1}{x^2 + 1} dx\]

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

Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]

 


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

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

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

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


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

\[\  ∫    x   \text{ e}^{x^2} dx\]

\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

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

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

`  ∫ \sqrt{"cosec x"- 1}  dx `

\[\int\frac{2x - 3}{x^2 + 6x + 13} dx\]

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

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

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

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

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

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

\[\int \log_{10} x\ dx\]

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

\[\int e^x \left( \cos x - \sin x \right) dx\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]

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

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

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

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

\[\int \text{cosec}^2 x \text{ cos}^2 \text{  2x  dx} \]

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

\[\int x\sqrt{2x + 3} \text{ dx }\]

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

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


\[\int\frac{6x + 5}{\sqrt{6 + x - 2 x^2}} \text{ dx}\]

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

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

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

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

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

Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×