English

∫ X 3 X − 2 D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int\left( \frac{x^3}{x - 2} \right)dx\]
\[ = \int\left( \frac{x^3 - 8 + 8}{x - 2} \right)dx\]
\[ = \int\left[ \frac{\left( x^3 - 2^3 \right)}{\left( x - 2 \right)} + \frac{8}{x - 2} \right]dx\]
\[ = \int\left[ \frac{\left( x - 2 \right)\left( x^2 + 2x + 4 \right)}{\left( x - 2 \right)} + \frac{8}{x - 2} \right]dx\]
\[ = \int\left( x^2 + 2x + 4 \right)dx + 8\int\frac{dx}{x - 2}\]
\[ = \frac{x^3}{3} + \frac{2 x^2}{2} + 4x + \text{8 ln} \left| x - 2 \right| + C\]
\[ = \frac{x^3}{3} + x^2 + 4x + \text{8 ln }\left| x - 2 \right| + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.04 [Page 30]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.04 | Q 2 | Page 30

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

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

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

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

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

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

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

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

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

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

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

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

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

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

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

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

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

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

\[\int {cosec}^3 x\ dx\]

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

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

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


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

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

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

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

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

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

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then


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

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

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

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

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

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

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

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


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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×