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Exercise: Standard Forms — Square Roots · Q27

Q.Evaluate ∫x+3x2+6x+13 dx\displaystyle\int \frac{x+3}{\sqrt{x^2+6x+13}}\,dx.

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The radicand's derivative is 2x+6=2(x+3)2x+6=2(x+3), exactly twice the numerator. Let u=x2+6x+13u=x^2+6x+13,

du=(2x+6) dx=2(x+3) dxdu=(2x+6)\,dx=2(x+3)\,dx, so (x+3) dx=12du(x+3)\,dx=\tfrac12du:

∫x+3x2+6x+13 dx=12∫u−1/2 du=12⋅2u+C=x2+6x+13+C.\int\frac{x+3}{\sqrt{x^2+6x+13}}\,dx = \frac12\int u^{-1/2}\,du = \frac12\cdot2\sqrt u+C = \sqrt{x^2+6x+13}+C. …

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