English

∫ 1 X 2 + 6 X + 13 D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{1}{x^2 + 6x + 13} dx\]
Sum
Advertisements

Solution

\[\int\frac{dx}{x^2 + 6x + 13}\]
\[ = \int\frac{dx}{x^2 + 2 \times x \times 3 + 9 - 9 + 13}\]
\[ = \int\frac{dx}{\left( x + 3 \right)^2 + 2^2}\]
\[ = \frac{1}{2} \tan^{- 1} \left( \frac{x + 3}{2} \right) + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.15 [Page 86]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.15 | Q 5 | Page 86

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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


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

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

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

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

\[\int\frac{a}{b + c e^x} dx\]

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

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

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

\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]

\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]

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

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

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

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

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

\[\int x \sin x \cos x\ dx\]

 


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

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

\[\int\cos\sqrt{x}\ dx\]

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

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

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

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

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

\[\int e^x \sec x \left( 1 + \tan x \right) dx\]

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

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

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

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

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

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

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

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

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

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

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

Find: `int (sin2x)/sqrt(9 - cos^4x) dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×