हिंदी

∫ X 6 + 1 X 2 + 1 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
योग
Advertisements

उत्तर

\[\int \left( \frac{x^6 + 1}{x^2 + 1} \right)dx\]
\[ = \int \left[ \frac{\left( x^2 \right)^3 + 1^3}{x^2 + 1} \right]\text{dx }A^3 + B^3 = \left( A + B \right) \left( A^2 - AB + B^2 \right)\]
\[ = \int\frac{\left( x^2 + 1 \right)\left( x^4 - x^2 + 1 \right)}{\left( x^2 + 1 \right)}dx\]
\[ = \int\left( x^4 - x^2 + 1 \right)dx\]
\[ = \int x^4 dx + \int x^2 dx + \int1dx\]
\[ = \frac{x^{4 + 1}}{4 + 1} - \frac{x^{2 + 1}}{2 + 1} + x + C\]
\[ = \frac{x^5}{5} - \frac{x^3}{3} + x + C\]

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 12 | पृष्ठ १४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]

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

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
 

\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]

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

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

\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]

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

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

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

\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

\[\int \cos^7 x \text{ dx  } \]

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

\[\int\frac{1}{x^2 - 10x + 34} dx\]

\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

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

\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

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

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

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

 


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

\[\int x^3 \tan^{- 1}\text{  x dx }\]

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

\[\int e^x \left( \cot x + \log \sin x \right) dx\]

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

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

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

 


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

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


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

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

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


\[\int x \sec^2 2x\ dx\]

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×