∫ 2 X + 1 √ X 2 + 2 X − 1 D X - Mathematics

प्रश्न

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

योग

उत्तर

\[\text{ Let I } = \int\frac{\left( 2x + 1 \right) dx}{\sqrt{x^2 + 2x - 1}}\]
\[ = \int\frac{\left( 2x + 2 - 1 \right) dx}{\sqrt{x^2 + 2x - 1}}\]
\[ = \int\frac{\left( 2x + 2 \right) dx}{\sqrt{x^2 + 2x - 1}} - \int\frac{dx}{\sqrt{x^2 + 2x - 1}}\]
\[ = \int\frac{\left( 2x + 2 \right) dx}{\sqrt{x^2 + 2x - 1}} - \int\frac{dx}{\sqrt{x^2 + 2x + 1 - 1 - 1}}\]
\[ = \int\frac{\left( 2x + 2 \right) dx}{\sqrt{x^2 + 2x - 1}} - \int\frac{dx}{\sqrt{\left( x + 1 \right)^2 - \left( \sqrt{2} \right)^2}}\]
\[\text{ let x}^2 + 2x - 1 = t\]
\[ \Rightarrow \left( 2x + 2 \right) dx = dt\]
\[I = \int\frac{dt}{\sqrt{t}} - \int\frac{dx}{\sqrt{\left( x + 1 \right)^2 - \left( \sqrt{2} \right)^2}}\]
\[ = 2\sqrt{t} - \text{ log} \left| x + 1 + \sqrt{\left( x + 1 \right)^2 - \left( \sqrt{2} \right)^2} \right| + C\]
\[ = 2\sqrt{x^2 + 2x - 1} - \text{ log }\left| x + 1 + \sqrt{x^2 + 2x - 1} \right| + C\]

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Indefinite Integral Problems

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Q 2 Q 1 Q 3

अध्याय 19: Indefinite Integrals - Exercise 19.21 [पृष्ठ ११०]

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आरडी शर्मा Mathematics [English] Class 12

अध्याय 19 Indefinite Integrals
Exercise 19.21 | Q 2 | पृष्ठ ११०

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