Evaluate ∫ x 2 − 1 x 2 d x

Given integral I = ∫ ( 1 + x 2 − 1 1 ) d x = ∫ d x + ∫ ( x − 1 ) ( x + 1 ) d x = x + 2 1 ∫ ( x − 1 1 − x + 1 1 ) d x = x + 2 1 lo g ( x + 1 x − 1 ) + c

Mathematics - Indefinite Integration Question with Solution | TestHub

MathematicsIndefinite IntegrationGeneralMedium2 minPYQ_2011

MathematicsMediumsingle choice

Evaluate $\int \frac{x ^{2}}{x ^{2} - 1} d x$

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Solution:

Given integral $I = \int (1 + \frac{1}{x ^{2} - 1}) d x$ $= \int d x + \int \frac{d x}{( x - 1 ) ( x + 1 )} = x + \frac{1}{2} \int (\frac{1}{x - 1} - \frac{1}{x + 1}) d x = x + \frac{1}{2} lo g (\frac{x - 1}{x + 1}) + c$

Stream:BITSATSubject:MathematicsTopic:Indefinite IntegrationSubtopic:General

⏱ 2mℹ️ Source: PYQ_2011

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