मराठी

∫ 2 X 4 + 7 X 3 + 6 X 2 X 2 + 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
बेरीज

उत्तर

\[\int\left( \frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} \right)dx\]
\[ = \int \frac{x^2 \left( 2 x^2 + 7x + 6 \right)}{x\left( x + 2 \right)}dx\]
\[ = \int\frac{x\left[ 2 x^2 + 4x + 3x + 6 \right]}{x + 2}dx\]
\[ = \int\frac{x\left( 2x\left( x + 2 \right) + 3\left( x + 2 \right) \right)}{\left( x + 2 \right)}dx\]

`= c x ( (2x+3)(x+2)) / (x+2) dx` 


\[ = \int\left( 2 x^2 + 3x \right)dx\]
` = 2   ∫  x^2 dx + 3∫   x    dx`
\[ = 2\left[ \frac{x^3}{3} \right] + 3\left[ \frac{x^2}{2} \right] + C\]
\[ = \frac{2}{3} x^3 + \frac{3}{2} x^2 + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.02 | Q 19 | पृष्ठ १५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

 
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec         } {x }- \cot x} dx\]

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

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

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

Integrate the following integrals:

\[\int\text { sin  x  cos  2x     sin 3x   dx}\]

` ∫ {"cosec"   x }/ { log  tan   x/2 ` dx 

\[\int \tan^{3/2} x \sec^2 \text{x dx}\]

\[\int \cot^6 x \text{ dx }\]

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

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

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

\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]

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

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

\[\int\frac{1}{1 - \cot x} dx\]

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

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

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

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

\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]

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

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

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

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

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

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

 


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

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

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

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

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

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

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

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

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×