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Solution (a) All the antiderivatives of $f(x)=x^{4}$ are $F(x) =\dfrac{x^{5}}{5}+C$, so $$\int x^{4}\, dx=\dfrac{x^{5}}{5}+C$$

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(b) All the antiderivatives of $f(x) =\sqrt{x} =x^{1/2}$ are $F(x) =\dfrac{x^{3/2}}{\dfrac{3}{2}}+C=\dfrac{2x^{3/2}}{3}+C$, so $$\int \sqrt{x} dx=\dfrac{2x^{3/2}}{3}+C$$

(c) No antiderivative in Table 1 corresponds to $f(x)=\dfrac{\sin x}{\cos ^{2}x}$, so we begin by using trigonometric identities to rewrite $\dfrac{\sin x}{\cos ^{2}x}$ in a form whose antiderivative is recognizable. $$\dfrac{\sin x}{\cos ^{2}x}=\dfrac{\sin x}{\cos x\cdot \cos x}=\dfrac{1}{\cos x}\cdot \dfrac{\sin x}{\cos x}=\sec x\tan x$$

Then $$\int \dfrac{\sin x}{\cos ^{2}x}\, dx=\int \sec x\tan x dx=\sec x+C$$