हिंदी

∫ X + 3 ( X + 1 ) 4 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.03 | Q 4 | पृष्ठ २३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

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

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

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

\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]

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

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

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

` ∫   tan   x   sec^4  x   dx  `


` ∫      tan^5    x   dx `


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

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

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

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

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

\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]

\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]

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

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

\[\int\frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \text{ dx }\]

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

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

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

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

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

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

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

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

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

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

\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

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

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

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

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


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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×