English

∫ ( Log X ) 2 ⋅ X D X - Mathematics

Advertisements
Advertisements

Question

\[\int \left( \log x \right)^2 \cdot x\ dx\]
Sum

Solution

\[\int \left( \log x \right)_{}^2 {x \cdot}    dx\]
`    "Taking  log x"^2" as the first function and x as the second function ". `
\[ = \left( \log x \right)^2 \int xdx - \int\left\{ \frac{d}{dx} \left( \log x \right)^2 \int x\ dx \right\}dx\]
\[ = \left( \log x \right)^2 \cdot \frac{x^2}{2} - \int\frac{\left( 2 \log x \right)}{x} \times \frac{x^2}{2} dx\]
\[ = \left( \log x \right)^2 \times \frac{x^2}{2} - \int x_{II} \log x_I dx\]
\[ = \left( \log x \right)^2 \times \frac{x^2}{2} - \left[ \log x \int x\ dx - \int\left\{ \frac{d}{dx}\left( \log x \right)\int x\ dx \right\}dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^2}{2} - \left[ \log x \cdot \frac{x^2}{2} - \int\frac{1}{x} \times \frac{x^2}{2}dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^2}{2} - \log x \cdot \frac{x^2}{2} + \frac{x^2}{4} + C\]
\[ = \frac{x^2}{2}\left[ \left( \log x \right)^2 - \log x + \frac{1}{2} \right] + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.25 [Page 133]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.25 | Q 21 | Page 133

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


\[\int \left( a \tan x + b \cot x \right)^2 dx\]

\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]

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

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

\[\int2x    \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]

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

` ∫  sec^6   x  tan    x   dx `

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

 


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

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

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


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

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

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

 
` ∫  x tan ^2 x dx 

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

\[\int\left( e^\text{log  x} + \sin x \right) \text{ cos x dx }\]


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

\[\int x \sin x \cos 2x\ dx\]

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

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

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

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

\[\int\sqrt{\cot \text{θ} d  } \text{ θ}\]

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

\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to

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

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

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

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

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

\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]

\[\int \sec^6 x\ dx\]

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

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

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×