3.3 Answer

$\begin{displaymath}\begin{array}{ll} u = \log{x}, & dv = x dx\begin{rawhtml} とおく... ...}{x}dx, & v = \int dv = \int x dx = \frac{x^2}{2} \end{array} \end{displaymath}$

$\displaystyle \int \underbrace{\log{x}}_{u} \underbrace{x dx}_{dv}$	$\displaystyle =$	$\displaystyle \underbrace{\frac{x^2}{2}}_{v} \underbrace{\log{x}}_{u} - \int \underbrace{\frac{x^2}{2}}_{v} \underbrace{\frac{1}{x} dx}_{du}$
	$\displaystyle =$	$\displaystyle \frac{x^2 \log{x}}{2} - \int \frac{x}{2}dx = \frac{x^2 \log{x}}{2} - \frac{x^2}{4} + c$

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

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

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

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

$\displaystyle \int x^2 e^{-x} dx = \frac{-2 - 2\,x - {x^2}}{{e^x}}$

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

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

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

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

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

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

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

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

$\displaystyle{I = \int e^{x} \sin{x} dx = -e^{x}\cos{x} + \int e^{x} \cos{x}dx }$ We use the integration by parts one more time. Set

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \int x^{2} \sin{x}dx = x^2 \sin{x} + 2 \int x\sin{x} dx$

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

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

$\displaystyle \int x^{3} \sin{x}dx = 6\,x\,\cos (x) - {x^3}\,\cos (x) - 6\,\sin (x) + 3\,{x^2}\,\sin (x)$

(k) To find $\int{x \sinh{x}}dx$ , one way to do is $\sinh{x} = \frac{e^{x} - e^{-x}}{2}$ or $(\sinh{x})' = \cosh{x}$ ．Set

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

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