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Worked solution

Use square roots of the standard-form constants

Integrate with respect to xx: (a) 110+x2\frac1{10+x^2}; (b) 12−x2\frac1{\sqrt{2-x^2}}.

CJC · Lecture · Self-Practice 4

(a)

Since 10=(10)210=(\sqrt{10})^2, ∫110+x2 dx=110tan⁡−1(x/10)+C\int\frac1{10+x^2}\,dx=\frac1{\sqrt{10}}\tan^{-1}(x/\sqrt{10})+C for all real xx.

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(b)

Since 2=(2)22=(\sqrt2)^2, ∫12−x2 dx=sin⁡−1(x/2)+C\int\frac1{\sqrt{2-x^2}}\,dx=\sin^{-1}(x/\sqrt2)+C for ∣x∣<2|x|<\sqrt2.

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Common mistakes

The printed answers use square root 10 and square root 2; native extraction can lose their radical signs.

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