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How would you solve the intigral from -1 to 2 for the equation (x/(x^2+1))dx?
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let u =x^2 +1 du = 2xdx rewrite the integral as: 1/2 Integral (2xdx/(x^2+1))
1/2 Integral (du/u) = 1/2 Ln|u| = 1/2 Ln(x^2+1) + C
Note you dont need abs value any more because x^2 is always positive and positive plus one is still positive!
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wait... is the anser 1/2ln5/2 or 1/2ln3?
oh sorry forgot to evaluate: 1/2 Ln(x^2+1) from -1 to 2: (1/2){Ln(2+1) - Ln(1+1)} = (1/2) {Ln3 - Ln2} = (1/2)Ln(3/2)
oh wait no
1/2ln5/2
2^2 is 4, not 2... (1/2){Ln(4+1) - Ln(1+1)} = (1/2) {Ln5 - Ln2} = (1/2)Ln(5/2)
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yep ok that helps alot ty
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