हिंदी

∫ X 2 + 5 X + 2 X + 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

\[\int\frac{\left( x^2 + 5x + 2 \right)}{\left( x + 2 \right)}dx\]
`=  ∫ x^2 / {x+2}  dx  + 5 ∫   {x   dx} / {x+2 } + 2 ∫  dx/{ x+2}`
\[ = \int\left( \frac{x^2 - 4 + 4}{x + 2} \right)dx + 5\int\left( \frac{x + 2 - 2}{x + 2} \right)dx + 2\int\frac{dx}{x + 2}\]
\[ = \int\frac{\left( x - 2 \right)\left( x + 2 \right)}{\left( x + 2 \right)}dx + \int\frac{4}{x + 2}dx + 5\int\left( 1 - \frac{2}{x + 2} \right)dx + 2\int\frac{dx}{x + 2}\]
\[ = \int\left( x - 2 \right) dx + 4\int\frac{dx}{x + 2} + 5\  ∫ dx - 10\int\frac{dx}{x + 2} + 2\int\frac{dx}{x + 2}\]
\[ = \int\left( x - 2 \right)dx - 4\int\frac{dx}{x + 2} + 5\  ∫  dx\]
\[ = \left( \frac{x^2}{2} - 2x \right) -\text{ 4  ln }\left| x + 2 \right| + 5x + C\]
\[ = \frac{x^2}{2} + 3x - \text{4  ln} \left| x + 2 \right| + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.04 [पृष्ठ ३०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.04 | Q 1 | पृष्ठ ३०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

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

\[\int \tan^2 \left( 2x - 3 \right) dx\]


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

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

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

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

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

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

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

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

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

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

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

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

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

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

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

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

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

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

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

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

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

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then


If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]


If `int(2x^(1/2))/(x^2)  dx = k  .  2^(1/x) + C`, then k is equal to ______.


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


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

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

\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]

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

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

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

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

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

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

\[\int\frac{x^2}{x^2 + 7x + 10} dx\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×