English

∫ 5 X + 3 √ X 2 + 4 X + 10 D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \text{ dx }\]
Sum
Advertisements

Solution

\[\text{ Let I }= \int\frac{\left( 5x + 3 \right) dx}{\sqrt{x^2 + 4x + 10}}\]
\[\text{ Consider, }\]
\[5x + 3 = A \frac{d}{dx} \left( x^2 + 4x + 10 \right) + B\]
\[ \Rightarrow 5x + 3 = A \left( 2x + 4 \right) + B\]
\[ \Rightarrow 5x + 3 = \left( 2A \right) x + 4A + B\]
\[\text{Equating Coefficients of like terms}\]
\[\text{ 2  A} = 5\]
\[ \Rightarrow A = \frac{5}{2}\]
\[\text{ And }\]
\[4A + B = 3\]
\[ \Rightarrow 4 \times \frac{5}{2} + B = 3\]
\[ \Rightarrow B = - 7\]
\[ \therefore I = \frac{5}{2}\int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 10}} - 7\int\frac{dx}{\sqrt{x^2 + 4x + 10}}\]
\[ = \frac{5}{2}\int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 10}} - 7\int\frac{dx}{\sqrt{x^2 + 4x + 4 - 4 + 10}}\]
\[ = \frac{5}{2}\int\frac{\left( 2x + 4 \right) dx}{\sqrt{x^2 + 4x + 10}} - 7\int\frac{dx}{\sqrt{\left( x + 2 \right)^2 + \left( \sqrt{6} \right)^2}}\]
\[\text{ Putting,} x^2 + 4x + 10 = t\]
\[ \Rightarrow \left( 2x + 4 \right) dx = dt\]
\[\text{ Then,} \]
\[I = \frac{5}{2}\int\frac{dt}{\sqrt{t}} - 7 \text{ log }\left| x + 2 + \sqrt{\left( x + 2 \right)^2 + 6} \right| + C\]
\[ = \frac{5}{2}\int t^{- \frac{1}{2}} dt - 7 \text{ log} \left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C\]
\[ = \frac{5}{2} \times 2\sqrt{t} - 7 \text{ log} \left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C\]
\[ = 5\sqrt{x^2 + 4x + 10} - 7 \text{ log }\left| x + 2 + \sqrt{x^2 + 4x + 10} \right| + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.21 [Page 111]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.21 | Q 17 | Page 111

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

Integrate the following integrals:

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

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

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

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

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

\[\int \cos^7 x \text{ dx  } \]

\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]

\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]

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

\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]

\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

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

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

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

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

\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×