मराठी

∫ 1 X S F R a C 1 3 ( X S F R a C 1 3 − 1 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

बेरीज

उत्तर

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


 ` ∫   1 / {x^{2/3} -   x^{1/3}}   ` dx
\[ \text{Let x }= t^3 \]
\[ \text{On differentiating both sides, we get}\]
\[ dx = 3 t^2 \text{  dt }\]
` ∴  I = ƒ {3 t^2}/ {( t^3 )^\{2/3} - ( t^3)^{1/3}}   dt  `


\[ = \int\frac{3 t^2}{t^2 - t}dt\]
\[ = 3\int\frac{t}{t - 1}dt\]
\[ = 3\int\frac{\left( t - 1 \right) + 1}{t - 1}dt\]
\[ = 3\int\left[ \left( 1 \right) + \frac{1}{t - 1} \right]dt\]
\[ = 3\left[ t + \text{ log }\left( t - 1 \right) \right] + c\]
` =  3 x ^{1/3 } + 3 log ( x^{1/3}  -1 ) + c `
Hence, ` ∫   1 /{x^{1/3} ( x^{1/3} -1)} dx = 3x^{1/3}   + 3  log  ( x^{1/3} -1 ) + c  `

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.10 [पृष्ठ ६५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.10 | Q 10 | पृष्ठ ६५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


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

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

\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]

\[\int\sqrt{1 + e^x} .  e^x dx\]

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

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

\[\int \cos^7 x \text{ dx  } \]

Evaluate the following integrals:

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

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

\[\int\frac{\cos x}{\sin^2 x + 4 \sin x + 5} dx\]

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

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

\[\int\frac{x + 2}{2 x^2 + 6x + 5}\text{  dx }\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{  dx }\]

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

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

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

\[\int\frac{1}{3 + 4 \cot x} dx\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

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

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

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

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

\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]

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

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

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

\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]

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

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

\[\int\frac{x^3}{x + 1}dx\] is equal to

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


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


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

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×