Question-172818

Question Number 172818 by Mikenice last updated on 01/Jul/22

Answered by FelipeLz last updated on 02/Jul/22

x^{4} + 2 x^{2} + 1 = {(x^{2} + 1)}^{2}

I = \int \frac{x^{3} + x^{2} + 2 x + 1}{x^{4} + 2 x^{2} + 1} d x = \int \frac{A}{x^{2} + 1} d x + \int \frac{B}{{(x^{2} + 1)}^{2}} d x

A (x^{2} + 1) + B = x^{3} + x^{2} + 2 x + 1

A (x^{2} + 1) + B = (x + 1) x^{2} + x + 1 + x

A (x^{2} + 1) + B = (x + 1) (x^{2} + 1) + x

A = x + 1 \land B = x

I = \int \frac{x + 1}{x^{2} + 1} d x + \int \frac{x}{{(x^{2} + 1)}^{2}} d x = \int \frac{x}{x^{2} + 1} d x + \int \frac{1}{x^{2} + 1} d x + \int \frac{x}{{(x^{2} + 1)}^{2}} d x

I = \frac{1}{2} [\ln (x^{2} + 1) - \frac{1}{x^{2} + 1}] + \tan^{- 1} (x) + c

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