I=∫_0^1x^2 2x^2+1dx=∫_0^1x^2+1 3x^2+1dx I=∫_0^1x^2+1x^2+1dx ∫_0^13x^2+1dx I=∫_0^1dx 3∫_0^11x^2+1dx I=[x]_0^1 3× [tan^ 1x]_0^1 I=1 3[tan ...

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Algebra of Limits

Solve ∫01 x2...

Question

Solve $\int_{0}^{1} \frac{x^{2} - 2}{x^{2} + 1} d x$

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Solution

$I = \int_{0}^{1} \frac{x^{2} - 2}{x^{2} + 1} d x = \int_{0}^{1} \frac{x^{2} + 1 - 3}{x^{2} + 1} d x$

$I = \int_{0}^{1} \frac{x^{2} + 1}{x^{2} + 1} d x - \int_{0}^{1} \frac{3}{x^{2} + 1} d x$

$I = \int_{0}^{1} d x - 3 \int_{0}^{1} \frac{1}{x^{2} + 1} d x$

$I = [x]_{0}^{1} - 3 \times [{tan}^{- 1} x]_{0}^{1}$

$I = 1 - 3 [{tan}^{- 1} 1 - {tan}^{- 1} 0]$

$I = 1 - 3 [\frac{π}{4} - 0]$

$I = 1 - \frac{3 π}{4}$

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