मराठी

∫ x 3 ( log x ) 2 dx - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int {x^3}_{II} \cdot \left( \log_I x \right)^2 \cdot dx\]
\[ = \left( \log x^2 \right)\int x^3 dx - \int\frac{2 \log x}{x} \times \frac{x^4}{4} \text{  dx} \]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2}\int \log_I x  \cdot {x^3}_{II} \text{  dx }\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2}\left[ \log x\int x^3 dx - \int\left\{ \frac{d}{dx}\left( \log x \right)\int x^3 dx \right\}dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2} \left[ \log x \cdot \frac{x^4}{4} - \int\frac{1}{x} \times \frac{x^4}{4}dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2} \left[ \log x \cdot \frac{x^4}{4} - \frac{1}{4}\int x^3 dx \right]\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{1}{2} \left[ \log x \cdot \frac{x^4}{4} - \frac{x^4}{16} \right] + C\]
\[ = \left( \log x \right)^2 \times \frac{x^4}{4} - \frac{\log x \cdot x^4}{8} + \frac{x^4}{32} + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Revision Excercise | Q 100 | पृष्ठ २०४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]

\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]

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

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

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


` ∫  tan^3    x   sec^2  x   dx  `

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

 


\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]

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

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

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

\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]

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


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

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

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

\[\int\frac{\log x}{x^n}\text{  dx }\]

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

 


\[\int {cosec}^3 x\ dx\]

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

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

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

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

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

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

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

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

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

\[\int \cos^3 (3x)\ dx\]

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

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

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×