3.3 解答

3.3

1.

(a)

$$\begin{array}{ll} u = \log{x}, & dv = x dxとおくと\\ du = \frac{1}{x}dx, & v = \int dv = \int x dx = \frac{x^2}{2} \end{array}$$

よって

$$\int \underbrace{\log{x}}_{u} \underbrace{x dx}_{dv} = \underbrace{\frac{x^2}{2}}_{v} \underbrace{\log{x}}_{u} - \int \underbrace{\frac{x^2}{2}}_{v} \underbrace{\frac{1}{x} dx}_{du}$$

$$= \frac{x^2 \log{x}}{2} - \int \frac{x}{2}dx = \frac{x^2 \log{x}}{2} - \frac{x^2}{4} + c$$

(b)

$$\begin{array}{ll} u = x^2, & dv = e^{-x} dxとおくと\\ du = 2x dx, & v = \int dv = \int e^{-x} dx = - e^{-x} \end{array}$$

よって

$$\int x^2 e^{-x} dx = -x^2 e^{-x} + 2 \int x e^{-x} dx$$

ここで部分積分をもう一度用いる.

$$\begin{array}{ll} u = x, & dv = e^{-x} dxとおくと\\ du = dx, & v = \int dv = \int e^{-x} dx = - e^{-x} \end{array}$$

これより

$$\int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + c$$

したがって

$${\int x^2 e^{-x} dx = -e^{x}(x^{2} + 2x + 2)}$$

(c)

$$\begin{array}{ll} u = (\log{x})^2, & dv = dxとおくと\\ du = 2\log{x}\frac{1}{x} dx, & v = \int dv = \int dx = x \end{array}$$

よって

$$\int (\log{x})^2 dx = x (\log{x})^2 - 2 \int \log{x} dx$$

ここで部分積分をもう一度用いる.

$$\begin{array}{ll} u = log{x}, & dv = dxとおくと\\ du = \frac{1}{x}dx, & v = \int dv = \int dx = x \end{array}$$

これより

$$\int \log{x} dx = x \log{x} - \int dx = x \log{x} - x + c$$

したがって $${\int (\log{x})^2 dx = x(\log{x})^2 - 2(x\log{x} - x) + c}$$

(d)

$t = x + 5$ とおくと $x = t-5$ ，$dt = dx$ よって

$$\int x(x+5)^{15} dx = \int (t - 5)t^{14} dt = \frac{1}{16}(x+5)^{16} - \frac{1}{3}(x+5)^{15} + c$$

(e)

$$\begin{array}{ll} u = x, & dv = \cos{x} dxとおくと\\ du = dx, & v = \int dv = \int \cos{x} dx = \sin{x} \end{array}$$

よって

$$\int x \cos{x}dx = x \sin{x} - \int \sin{x} dx = x \sin{x} + \cos{x} +c$$

(f)

$I = \int e^{x} \sin{x}dx$ とおいて部分積分を行う

$$\begin{array}{ll} u = e^{x}, & dv = \sin{x} dxとおくと\\ du = e^{x}dx, & v = \int dv = \int \sin{x} dx = -\cos{x} \end{array}$$

より

$${I = \int e^{x} \sin{x} dx = -e^{x}\cos{x} + \int e^{x} \cos{x}dx }$$ ここで部分積分をもう一度行うと

$$\begin{array}{ll} u = e^{x}, & dv = \cos{x} dxとおくと\\ du = e^{x}dx, & v = \int dv = \int \cos{x} dx = \sin{x} \end{array}$$

より

$$I = \int e^{x} \sin{x} dx = -e^{x}\cos{x} + \int e^{x} \cos{x}dx$$

$$= -e^{x}\cos{x} + e^{x}\sin{x} + \int e^{x} \sin{x}dx$$

したがって求める積分$I$ は

$$I = -\frac{e^{x}}{2}(\cos{x} - \sin{x})$$

(g)

$$\begin{array}{ll} u = \log{(1+x^2)}, & dv = dxとおくと\\ du = \frac{2x}{1 + x^2}dx, & v = \int dv = \int dx = x \end{array}$$

より

$$\int \log{(1 + x^2)}dx = x \log{(1+x^2)} - 2\int \frac{x^2}{1+x^2}dx$$

ここで

$$\int \frac{x^2}{1+x^2}dx = \int \frac{1 + x^2 - 1}{1 + x^2} dx = \int (1 - \frac{1}{1 + x^2} dx = x - \tan^{-1}{x}$$

に注意すると

$$\int \log{(1 + x^2)}dx = x \log{(1+x^2)} - 2(x - \tan^{-1}{x}) + c$$

(h)

$$\begin{array}{ll} u = \tan^{-1}{x}, & dv = x dxとおくと\\ du = \... ...x^2}dx, & v = \int dv = \int x dx = \frac{x^2}{2} \end{array}$$

より

$$\int \log{(1 + x^2)}dx = \frac{x^2 \tan^{-1}{x}}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2}dx$$

ここで

$$\int \frac{x^2}{1+x^2}dx = \int \frac{1 + x^2 - 1}{1 + x^2} dx = \int (1 - \frac{1}{1 + x^2} dx = x - \tan^{-1}{x}$$

に注意すると

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

(i)

$$\begin{array}{ll} u = \log{x}, & dv = x^n dxとおくと\\ du = \fra... ...& v = \int dv = \int x^n dx = \frac{x^{n+1}}{n+1} \end{array}$$

より

$$\int x^{n} \log{x}dx = \frac{x^{n+1} \log{x}}{n+1} - \frac{1}{n+1} \int x^{n} dx$$

$$= \frac{x^{n+1} \log{x}}{n+1} - \frac{x^{n+1}}{n+1} + c$$

(j)

$$\begin{array}{ll} u = x^{3}, & dv = \sin{x} dxとおくと\\ du = 3x^2 dx, & v = \int dv = \int \sin{x} dx = - \cos{x} \end{array}$$

より

$${\int x^{3} \sin{x}dx = -x^3 \cos{x} + 3 \int x^{2}\cos{x} dx}$$

$\int x^{2}\cos{x} dx$ も同様に積分すると

$$\begin{array}{ll} u = x^{2}, & dv = \cos{x} dxとおくと\\ du = 2x dx, & v = \int dv = \int \cos{x} dx = \sin{x} \end{array}$$

より

$${\int x^{2} \cos{x}dx = x^2 \sin{x} - 2 \int x\sin{x} dx}$$

$\int x \sin{x} dx$ も同様に積分すると

$$\begin{array}{ll} u = x, & dv = \sin{x} dxとおくと\\ du = dx, & v = \int dv = \int \sin{x} dx = -\cos{x} \end{array}$$

より

$${\int x \sin{x}dx = -x \cos{x} + \int \cos{x} dx = -x \cos{x} + \sin{x} + c}$$

したがって

$${-x^{3}\cos{x} + 3x^{2}\sin{x} + 6x\cos{x} - 6\sin{x}}$$

(k) $\int{x \sinh{x}}dx$の積分は $\sinh{x} = \frac{e^{x} - e^{-x}}{2}$と書き直して求めるか， $(\sinh{x})' = \cosh{x}$を用いて求める．

$$\begin{array}{ll} u = x, & dv = \sinh{x} dxとおくと\\ du = dx, & v = \int dv = \int \sinh{x} dx = \cosh{x} \end{array}$$

より

$${\int x \sin{x}dx = x \cosh{x} - \int \cosh{x} dx = x \cosh{x} - \sinh{x} + c}$$