English

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

Advertisements
Advertisements

Question

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

Solution

We have,
\[I = \int \frac{\left( 2x + 1 \right)dx}{\left( x - 2 \right) \left( x - 3 \right)}\]
\[\text{Let }\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{A}{x - 2} + \frac{B}{x - 3}\]
\[ \Rightarrow \frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{A\left( x - 3 \right) + B\left( x - 2 \right)}{\left( x - 2 \right) \left( x - 3 \right)}\]
\[ \Rightarrow 2x + 1 = A\left( x - 3 \right) + B\left( x - 2 \right)\]
\[\text{Putting }x - 3 = 0\]
\[ \Rightarrow x = 3\]
\[ \therefore 7 = A \times 0 + B \times \left( 3 - 2 \right)\]
\[ \Rightarrow B = 7\]
\[\text{Putting }x - 2 = 0\]
\[ \Rightarrow x = 2\]
\[ \therefore 5 = A\left( - 1 \right)\]
\[ \Rightarrow A = - 5\]
\[ \therefore I = - 5\int\frac{dx}{x - 2} + 7\int\frac{dx}{x - 3}\]
\[ = - 5 \log \left| x - 2 \right| + 7 \log \left| x - 3 \right| + C\]
\[ = \log \left| x - 3 \right|^7 - \log \left| x - 2 \right|^5 + C\]
\[ = \log \left| \frac{\left( x - 3 \right)^7}{\left( x - 2 \right)^5} \right| + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.30 [Page 177]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.30 | Q 52 | Page 177

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]

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

`∫     cos ^4  2x   dx `


` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


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

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

\[\ \int\ x \left( 1 - x \right)^{23} dx\]

 


\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

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

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

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

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

\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

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

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

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

 
` ∫  x tan ^2 x dx 

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

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

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

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

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

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

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

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

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


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

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

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

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

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

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


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


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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×