मराठी

∫ X 2 X 6 + a 6 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{x^2}{x^6 + a^6} dx\]
बेरीज

उत्तर

\[\int\frac{x^2}{x^6 + a^6}dx\]
\[ \Rightarrow \int\frac{x^2 dx}{\left( x^3 \right)^2 + \left( a^3 \right)^2}\]
\[\text{ let } x^3 = t\]
\[ \Rightarrow 3 x^2 dx = dt\]
\[ \Rightarrow x^2 dx = \frac{dt}{3}\]
\[Now, \int\frac{x^2}{x^6 + a^6}dx\]
\[ = \frac{1}{3}\int\frac{dt}{t^2 + \left( a^3 \right)^2}\]
\[ = \frac{1}{3 a^3} \tan^{- 1} \left( \frac{t}{a^3} \right) + C\]
\[ = \frac{1}{3 a^3} \tan-^1 \left( \frac{x^3}{a^3} \right) + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.16 [पृष्ठ ९०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.16 | Q 10 | पृष्ठ ९०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

\[\int\text{sin mx }\text{cos nx dx m }\neq n\]

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

\[\int\sqrt {e^x- 1}  \text{dx}\] 

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

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

\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]

\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]

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

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

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


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

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

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

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

\[\int\frac{1}{p + q \tan x} \text{ dx  }\]

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

 
` ∫  x tan ^2 x dx 

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

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

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

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

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

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

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

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

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

Evaluate the following integral:

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

\[\int\sqrt{\cot \text{θ} d  } \text{ θ}\]

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

` \int \text{ x} \text{ sec x}^2 \text{  dx  is  equal  to }`

 


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

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

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

\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×