3.1 解答

3.1

(a)

$\displaystyle{\int (3x^{-3} + 4x^{5}) dx = 3(-\frac{1}{2}x^{-2}) + \frac{2}{3}x^6 = -\frac{3}{2}x^{-2} + \frac{2}{3}x^{6} + c }$

(b)

$\displaystyle{\int (t^3 + \frac{1}{t^2 + 1}) dt = \frac{t^4}{4} + \tan^{-1}(t) + c}$

(c)

$\displaystyle \int -2 \tan^{2}{x} dx$	$\displaystyle =$	$\displaystyle -2 \int \frac{\sin^{2}{x}}{\cos^{2}{x}} dx = -2 \int \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}dx$
	$\displaystyle =$	$\displaystyle -2 \int (\sec^{2}{x} - 1)dx = -2(\tan{x} - x) + c$

(d)

$\displaystyle \int \frac{x^3 + 1}{\sqrt{x}} dx$	$\displaystyle =$	$\displaystyle \int x^{\frac{5}{2}} + x^{-\frac{1}{2}} dx$
	$\displaystyle =$	$\displaystyle \frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{1}{2}} + c$

(e)

$\displaystyle \int \frac{x^2 + 5}{x^2 + 4} dx$	$\displaystyle =$	$\displaystyle \int 1 + \frac{1}{x^2 + 4} dx =$
	$\displaystyle =$	$\displaystyle x + \frac{1}{2}\tan^{-1}{\frac{x}{2}} + c$

(f)

$\displaystyle \int \frac{1}{\sqrt{4 - t^2}} dt$	$\displaystyle =$	$\displaystyle \int \frac{1}{\sqrt{2^2 - t^2}} dt =$
	$\displaystyle =$	$\displaystyle \sin^{-1}{(\frac{t}{2})} + c$

(g)

$\displaystyle \int \cos^{2}{x} dx$	$\displaystyle =$	$\displaystyle \int \frac{1 + \cos{2x}}{2} dx =$
	$\displaystyle =$	$\displaystyle \frac{x}{2} + \frac{\sin{2x}}{4} + c$

(h)

$\displaystyle{\int \frac{1}{\sqrt{t^2 + 4}} dt = \log{\vert t + \sqrt{t^2 + 4}\vert} + c}$

(i)

$\displaystyle{\frac{1}{4 - x^2} dx = \frac{1}{4}\log{\left\vert\frac{2+x}{2-x}\right\vert} + c}$