मराठी

∫ 3 X 5 1 + X 12 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{3 x^5}{1 + x^{12}} dx\]
बेरीज
Advertisements

उत्तर

\[\int\frac{3 x^5}{1 + x^{12}}dx\]
\[\text{ let } x^6 = t\]
\[ \Rightarrow 6 x^5 dx = dt\]
\[ \Rightarrow x^5 dx = \frac{dt}{6}\]
\[Now, \int\frac{3 x^5}{1 + x^{12}}dx\]
\[ = \frac{3}{6}\int\frac{dt}{1 + t^2}\]
\[ = \frac{1}{2} \tan^{- 1} \left( t \right) + C\]

\[= \frac{1}{2} \tan^{- 1} \left( x^6 \right) + C\]
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 8 | पृष्ठ ९०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

\[\int \left( e^x + 1 \right)^2 e^x dx\]

`  ∫  sin 4x cos  7x  dx  `

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

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

 ` ∫   1 /{x^{1/3} ( x^{1/3} -1)}   ` dx


\[\int \sin^7 x  \text{ dx }\]

\[\int \sin^3 x \cos^5 x \text{ dx  }\]

` = ∫1/{sin^3 x cos^ 2x} dx`


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

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

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

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

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

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

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

\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]

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

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

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

\[\int\frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \text{ dx }\]

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

\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to

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

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

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


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


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

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

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

\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]

\[\int\sqrt{\frac{a + x}{x}}dx\]
 

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

\[\int \log_{10} x\ dx\]

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

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

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

\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×