English

∫ X + 1 X 2 + X + 3 D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int\frac{\left( x + 1 \right) dx}{x^2 + x + 3}\]
\[x + 1 = \frac{Ad}{dx}\left( x^2 + x + 3 \right) + B\]
\[x + 1 = A \left( 2x + 1 \right) + B\]
\[x + 1 = \text{ 2 Ax + A + B }\]

Comparing Coefficients of like powers of x

\[2A = 1\]
\[A = \frac{1}{2}\]
\[A + B = 1\]
\[\frac{1}{2} + B = 1\]
\[B = \frac{1}{2}\]
\[\left( x + 1 \right) = \frac{1}{2} \left( 2x + 1 \right) + \frac{1}{2}\]

\[Now, \int\frac{\left( x + 1 \right) dx}{x^2 + x + 3}\]
\[ = \int\frac{\frac{1}{2} \left( 2x + 1 \right)dx}{x^2 + x + 3} + \frac{1}{2}\int\frac{dx}{x^2 + x + 3}\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 1 \right)dx}{x^2 + x + 3} + \frac{1}{2}\int\frac{dx}{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 3}\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 1 \right)dx}{x^2 + x + 3} + \frac{1}{2}\int\frac{dx}{\left( x + \frac{1}{2} \right)^2 + 3 - \frac{1}{4}}\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 1 \right) dx}{x^2 + x + 3} + \frac{1}{2}\int\frac{dx}{\left( x + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{11}}{2} \right)^2}\]
\[ = \frac{1}{2} \text{ log }\left| x^2 + x + 3 \right| + \frac{1}{2} \times \frac{2}{\sqrt{11}} \text{ tan}^{- 1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{11}}{2}} \right) + C\]
\[ = \frac{1}{2} \text{ log }\left| x^2 + x + 3 \right| + \frac{1}{\sqrt{11}} \text{ tan}^{- 1} \left( \frac{2x + 1}{\sqrt{11}} \right) + C\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.19 [Page 104]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.19 | Q 2 | Page 104

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

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

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

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


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

\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]

\[\int2x    \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]

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

` ∫  tan^5 x   sec ^4 x   dx `

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

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

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


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


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

\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]

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

` ∫    sin x log  (\text{ cos x ) } dx  `

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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


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

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

Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×