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integration help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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∫(x^2+2x)dx/(x+1)^2

asked Jun 21, 2013 in CALCULUS by linda Scholar

1 Answer

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∫(x^2+2x)dx/(x+1)^2

= ∫(x^2 + 2x + 1 - 1)dx/(x+1)^2

= ∫((x + 1)^2 - 1)dx/(x+1)^2

= ∫((x + 1)^2) / ((x + 1)^2)dx - dx/(x+1)^2

Cancel common terms.

= ∫dx - dx(x+1)^(-2)

Formula : ∫x^n = (x^(n+1))/(n+1)

= x - {(x+1)^(-2+1)} / (-2+1)

= x - {(x+1)^(-1)} / (-1)

= x + {(x+1)^(-1)}

= x + {1/(x+1)}

answered Jun 21, 2013 by anonymous

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