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

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प्रश्न

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

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

Putting x + 1 = t
⇒ x = t – 1
& dx = dt

\[\therefore I = \int\left[ \frac{\left( t - 1 \right)^2 + 3 \left( t - 1 \right) - 1}{t^2} \right]dt\]
\[ = \int \left( \frac{t^2 - 2t + 1 + 3t - 3 - 1}{t^2} \right)dt\]
\[ = \int\left( \frac{t^2 + t - 3}{t^2} \right)dt\]
\[ = \int\left( 1 + \frac{1}{t} - 3 t^{- 2} \right)dt\]
\[ = t + \text{log} \left| t \right| - 3\left( \frac{t^{- 2 + 1}}{- 2 + 1} \right) + C\]
\[ = t + \text{log}\left| t \right| + \frac{3}{t} + C\]
\[ = x + 1 + \text{log} \left| x + 1 \right| + \frac{3}{x + 1} + C \left[ \because t = x + 1 \right]\]

Let C + 1 = C′

\[= x + \text{log} \left( x + 1 \right) + \frac{3}{x + 1} + C\prime\]
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अध्याय 19: Indefinite Integrals - Exercise 19.04 [पृष्ठ ३०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.04 | Q 5 | पृष्ठ ३०

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