मराठी

∫ X 3 X − 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{x^3}{x - 2} dx\]
बेरीज

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.04 [पृष्ठ ३०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.04 | Q 2 | पृष्ठ ३०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\] 

` ∫ {"cosec"   x }/ { log  tan   x/2 ` dx 

` ∫  tan 2x tan 3x  tan 5x    dx  `

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

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

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

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

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

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

 


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

\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

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

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

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

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

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

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

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

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

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

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

Evaluate the following integral:

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

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

Write a value of

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

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

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

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

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

\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]

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

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

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

\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]


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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×