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What is the solution of the below integral:

D e a r s t u d e n t, \int \frac{x^{2} - 4}{x + 1} d x L e t u = x + 1 S o, d u = d x x^{2} = {(u - 1)}^{2} \int \frac{{(u - 1)}^{2} - 4}{u} d u \int \frac{u^{2} - 2 u + 1 - 4}{u} d u \int \frac{u^{2} - 2 u - 3}{u} d u \int (\frac{u^{2}}{u} - \frac{2 u}{u} - \frac{3}{u}) d u \int \frac{u^{2}}{u} d u - \int \frac{2 u}{u} d u - \int \frac{3}{u} d u \int u d u - \int 2 d u - \int \frac{3}{u} d u \frac{u^{2}}{2} - 2 u - 3 \ln u \frac{{(x + 1)}^{2}}{2} - 2 (x + 1) - 3 \ln (x + 1) 1 . \int x^{n} d x = \frac{x^{n + 1}}{n + 1} 2 . \int \frac{1}{x} d x = \ln x R e g a r d s

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