हिंदी

∫ ( X 2 + 1 ) ( X 2 + 2 ) ( X 2 + 3 ) ( X 2 + 4 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

योग
Advertisements

उत्तर

We have,
\[I = \int \frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)}\]
\[\text{Putting }x^2 = t\]
Then,
\[\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} = \frac{\left( t + 1 \right) \left( t + 2 \right)}{\left( t + 3 \right) \left( t + 4 \right)} = \frac{t^2 + 3t + 2}{t^2 + 7t + 12}\]

Degree of numerator is equal to degree of denominator.

We divide numerator by denominator.

\[\therefore \frac{t^2 + 3t + 2}{t^2 + 7t + 12} = 1 - \left( \frac{4t + 10}{t^2 + 7t + 12} \right)\]
\[ \Rightarrow \frac{t^2 + 3t + 2}{t^2 + 7t + 12} = 1 - \frac{4t + 10}{\left( t + 3 \right) \left( t + 4 \right)} ............. \left( 1 \right)\]
\[\text{Let }\frac{4t + 10}{\left( t + 3 \right) \left( t + 4 \right)} = \frac{A}{t + 3} + \frac{B}{t + 4}\]
\[ \Rightarrow \frac{4t + 10}{\left( t + 3 \right) \left( t + 4 \right)} = \frac{A\left( t + 4 \right) + B\left( t + 3 \right)}{\left( t + 3 \right) \left( t + 4 \right)}\]
\[ \Rightarrow 4t + 10 = A\left( t + 4 \right) + B\left( t + 3 \right)\]
\[\text{Putting t + 4 = 0}\]
\[ \Rightarrow t = - 4\]
\[ \therefore - 16 + 10 = B\left( - 1 \right)\]
\[ \Rightarrow B = 6\]
\[\text{Putting t + 3 = 0}\]
\[ \Rightarrow t = - 3\]
\[ \therefore - 12 + 10 = A\left( - 3 + 4 \right)\]
\[ \Rightarrow A = - 2\]
\[ \therefore \frac{4t + 10}{\left( t + 3 \right) \left( t + 4 \right)} = \frac{- 2}{t + 3} + \frac{6}{t + 4} ................ \left( 2 \right)\]
From (1) and (2)
\[\frac{t^2 + 3t + 2}{t^2 + 7t + 12} = 1 + \frac{2}{t + 3} - \frac{6}{t + 4}\]
\[ \therefore \int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)dx}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} = \int dx + 2\int\frac{dx}{x^2 + \left( \sqrt{3} \right)^2} - 6\int\frac{dx}{x^2 + 2^2}\]
\[ = x + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{x}{\sqrt{3}} \right) - \frac{6}{2} \tan^{- 1} \left( \frac{x}{2} \right) + C\]
\[ = x + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{x}{\sqrt{3}} \right) - 3 \tan^{- 1} \left( \frac{x}{2} \right) + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.30 [पृष्ठ १७८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.30 | Q 63 | पृष्ठ १७८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

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

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


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

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

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

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

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

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

\[\int \cos^7 x \text{ dx  } \]

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

\[\int\frac{\cos x}{\sin^2 x + 4 \sin x + 5} dx\]

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

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

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


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

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

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

`int 1/(cos x - sin x)dx`

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

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

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

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

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

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


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


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

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

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

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

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

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

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

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

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

\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
 

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×