English

∫ 1 X ( X 6 + 1 ) D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

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


\[ = \int\frac{x^5 dx}{x^6 \left( x^6 + 1 \right)}\]
\[\text{ let }x^6 = t\]
\[ \Rightarrow 6 x^5 dx = dt\]


\[ \Rightarrow x^5 dx = \frac{dt}{6}\]
\[Now, \int\frac{dx}{x^6 \left( x^6 + 1 \right)}\]
\[ = \frac{1}{6}\int\frac{dt}{t\left( t + 1 \right)}\]
\[ = \frac{1}{6}\int\frac{dt}{t^2 + t}\]
\[ = \frac{1}{6}\int\frac{dt}{t^2 + t + \frac{1}{4} - \frac{1}{4}}\]
\[ = \frac{1}{6}\int\frac{dt}{\left( t + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\]
\[ = \frac{1}{6} \times \frac{1}{2 \times \frac{1}{2}} \text{ log  }\left| \frac{t + \frac{1}{2} - \frac{1}{2}}{t + \frac{1}{2} + \frac{1}{2}} \right| + C\]
\[ = \frac{1}{6} \text{ log }  \left| \frac{t}{t + 1} \right| + C\]
\[ = \frac{1}{6} \text{ log  }\left| \frac{x^6}{x^6 + 1} \right| + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.16 [Page 90]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.16 | Q 11 | Page 90

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

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

` ∫  1/ {1+ cos   3x}  ` dx


\[\int \tan^2 \left( 2x - 3 \right) dx\]


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

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]

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

\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]

\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]

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

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

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

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

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

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

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

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

 


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

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

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

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

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

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

Write a value of

\[\int e^{3 \text{ log x}} x^4\text{ dx}\]

Write the anti-derivative of  \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]


\[\int\frac{1}{7 + 5 \cos x} dx =\]

\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]

\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 


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

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

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


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

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


\[\int \sec^6 x\ dx\]

\[\int \tan^3 x\ \sec^4 x\ dx\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×