हिंदी

∫ ( 2 X + 5 X − 1 X 1 / 3 ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\left( 2^x + \frac{5}{x} - \frac{1}{x^{1/3}} \right)dx\]
योग
Advertisements

उत्तर

\[\int\left( 2^x + \frac{5}{x} - \frac{1}{x^\frac{1}{3}} \right)dx\]
\[ = \int 2^x dx + 5 \int\frac{dx}{x} - \int\frac{dx}{x^\frac{1}{3}}\]
\[ = \int 2^x dx + 5 \int\frac{dx}{x} - \int x^{- \frac{1}{3}} dx\]
\[ = \frac{2^x}{\ln 2} + 5 \ln x - \left[ \frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1} \right] + C\]
\[ = \frac{2^x}{\ln 2} + 5 \ln x - \frac{3}{2} x^\frac{2}{3} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.02 | Q 2 | पृष्ठ १४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]

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

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

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

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

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

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

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

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

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

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

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

\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]

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

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

\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]

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

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

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

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

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

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

\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{  dx }\]

 


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

\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]

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

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

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

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

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

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


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

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

\[\int x \sin^5 x^2 \cos x^2 dx\]

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

\[\int {cosec}^4 2x\ dx\]


Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×