मराठी

∫ 1 x √ 1 + x 3 dx - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{1}{x\sqrt{1 + x^3}} \text{ dx}\]
बेरीज
Advertisements

उत्तर

\[\text{We have}, \]
\[I = \int\frac{dx}{x \sqrt{1 + x^3}}\]
\[ = \int\frac{x^2 dx}{x^3 \sqrt{1 + x^3}}\]
\[\text{ putting  x}^3 = t\]
\[ \Rightarrow \text{  3 x}^2 \text{ dx }= dt\]
\[ \Rightarrow x^2 dx = \frac{dt}{3}\]
\[ \therefore I = \frac{1}{3}\int\frac{dt}{t\sqrt{1 + t}}\]
\[\text{ let 1 + t = p}^2 \]
\[ \Rightarrow \text{ dt = 2p dp}\]
\[I = \frac{1}{3}\int\frac{\text{ 2p dp}}{\left( p^2 - 1 \right) \times p}\]
\[ = \frac{2}{3}\int\frac{dp}{p^2 - 1}\]
\[ = \frac{2}{3} \times \frac{1}{2} \text{ log} \left| \frac{p - 1}{p + 1} \right| + C\]
\[ = \frac{1}{3}\text{ log }\left| \frac{p - 1}{p + 1} \right| + C\]
\[ = \frac{1}{3}\text{ log} \left| \frac{\sqrt{1 + t} - 1}{\sqrt{1 + t} + 1} \right| + C\]
\[ = \frac{1}{3}\text{ log } \left| \frac{\sqrt{1 + x^3} - 1}{\sqrt{1 + x^3} + 1} \right| + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Revision Excercise | Q 106 | पृष्ठ २०४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]

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

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

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

\[\int \cos^2 \text{nx dx}\]

` ∫   cos  3x   cos  4x` dx  

Integrate the following integrals:

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

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

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

` ∫    x   {tan^{- 1} x^2}/{1 + x^4} dx`

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

\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]

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

\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]

\[\int\frac{1}{5 + 4 \cos x} dx\]

\[\int\frac{\log x}{x^n}\text{  dx }\]

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

\[\int x \sin x \cos 2x\ dx\]

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

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

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

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

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

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

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

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

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

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

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


\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{  dx }\]

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

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

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

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

\[\int \tan^5 x\ \sec^3 x\ dx\]

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

\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×