हिंदी

∫ X + 3 ( X + 1 ) 4 D X

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प्रश्न

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

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

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 18 Indefinite Integrals
Exercise 19.03 | Q 4 | पृष्ठ २३
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