71. $\int 1/(cbrt(x)+1)dx$

$\begin{aligned} &\int \frac{1}{\sqrt[3]{x}+1}dx \\ &(u=\sqrt[3]{x}, du=\frac{1}{3\sqrt[3]{x^2}}dx=\frac{1}{3}x^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u+1}du=3\int \frac{u^2-1+1}{u+1}du\\ &=3\int \frac{(u-1)(u+1)}{u+1}du+3\int\frac{1}{u+1}du \\ &=3(\frac{1}{2}u^2-u)+3ln|u+1|\\ &=\frac{3}{2}u^2-3u+3ln|u+1|+C\\ &=\frac{3}{2}\sqrt[3]{x}^2-3\sqrt[3]{x}+3ln|\sqrt[3]{x}+1|+C\\ \end{aligned}$
Alt.
$(u=\sqrt[3]{x}+1, x=(u-1)^3, dx=3(u-1)^2du)\\ =\int \frac{1}{u} 3(u-1)^2du\\ =3\int \frac{u^2-2u+1}{u} du=3\int u-2+(1/u) du\\ =3(\frac{1}{2}u^2-2u+ln|u|)\\ =\frac{3}{2}(\sqrt[3]{x}+1)^2-6(\sqrt[3]{x}+1)+3ln|\sqrt[3]{x}+1|+C$

72. $\int \frac{1}{cbrt(x + 1)}dx$

$\begin{aligned} &\int \frac{1}{\sqrt[3]{x+1}}dx \\ &(u=\sqrt[3]{x+1}, du=\frac{1}{3\sqrt[3]{(x+1)^2}}dx=\frac{1}{3}(x+1)^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u}du=3\int u du \\ &=\frac{3}{2}u^2+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}+C \end{aligned}$
Alt.
(u=x+1)
$\begin{aligned} &\int \frac{1}{\sqrt[3]{x+1}}dx \\ &=\int u^{-\frac{1}{3}}du \\ &=\frac{3}{2}u^2+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}+C \end{aligned}$

73. $\int (sin(x)+cos(x))^2 dx$

$\begin{aligned} &\int (sin(x)+cos(x))^2dx \\ &=\int 1+2sin(x)cos(x)dx=x+\int sin(2x) dx\\ &=x-\frac{1}{2}cos(2x)+C \end{aligned}$

74. $\int 2xln(1+x) dx$

$\begin{aligned} &\int 2xln(1+x) dx =2\int xln(1+x) dx\\ & \text{(ln쪽을 미분하는 방향으로 부분적분)}\\ &=2( ln(1+x)\frac{1}{2}x^2-\int \frac{1}{1+x}\frac{1}{2}x^2dx )\\ &= ln(1+x)x^2-\int \frac{x^2}{1+x}dx = ln(1+x)x^2-\int \frac{x^2-1+1}{1+x}dx\\ &= ln(1+x)x^2-\int x-1+\frac{1}{1+x}dx\\ &= ln(1+x)x^2-\frac{1}{2}x^2+x-ln(1+x)+C\\ \end{aligned}$

75. $\int 1/(x(1+sin^2ln(x)))dx$

$\begin{aligned} &\int \frac{1}{x(1+sin^2ln(x))} dx \\ &( u=ln(x), du=\frac{1}{x}dx : 1/x과 ln(x)에서 치환 착안.)\\ &=\int \frac{du}{1+sin^2u} : (sin에 대해 치환, 변환(반각)해보고, cos으로 나눠도 본다.\\ & sec, tan을 만들 수 있다는 것을 착안.)\\ &=\int \frac{sec^2u}{sec^2u+tan^2u}du\\ &=\int \frac{sec^2u}{1+2tan^2u}du, (t=tanu, dt=sec^2udu)\\ &=\int \frac{dt}{1+2t^2}, (t=tanu, dt=sec^2udu)\\ &=\frac{1}{\sqrt{2}}arctan {\sqrt{2}t}=\frac{1}{\sqrt{2}}arctan({\sqrt{2} tan(ln(x))})+C \end{aligned}$

76. $\int sqrt((1-x)/(1+x))dx$

Try
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(\text{전체치환? 일부치환. 분자분모곱, pm 등})\\ &( u = \sqrt{\frac{1-x}{1+x}}, du=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{1-x+1+x}{1-2x+x^2}dx)\\ &( du = \frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{2}{(1-x)^2}dx=\frac{1}{u(1-x)^2}dx)\\ &(u^2=\frac{1-x}{1+x}, xu^2+x=1-u^2, x=\frac{1-u^2}{1+u^2} )\\ &=\int u^2(1-x)^2 du =\int u^2(1-\frac{1-u^2}{1+u^2})^2du\\ &=\int u^2(\frac{2u^2}{1+u^2})^2 du = \int 4u^6(\frac{1}{1+u^2})^2du\\ \end{aligned}$
Try
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(u=\frac{1-x}{1+x}, du=\frac{-1-x-(1-x)}{1+2x+x^2}dx=\frac{-2}{(1+x)^2}dx)\\ &( xu+u=1-x, x(u+1)=1-u, x=\frac{1-u}{1+u})\\ &=-\frac{1}{2}\int \sqrt{u} (1+x^2) du = -\frac{1}{2}\int \sqrt{u} (1+(\frac{1-2u+u^2}{1+2u+u^2})^2) du \\ &=-\frac{1}{2}\int \sqrt{u} (\frac{2+2u^2}{1+2u+u^2})^2 du \\ &=-2\int\sqrt{u}\frac{(1+u^2)^2}{(1+u)^4}du \end{aligned}$
Solve
$\begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx =\int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} dx\\ &=\int \sqrt{\frac{(1-x)^2}{1-x^2}} dx =\int \frac{1-x}{\sqrt{1-x^2}}dx\\ &=\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{x}{\sqrt{1-x^2}}dx\\ & (u=1-x^2, du=-2xdx)\\ &=\sin^{-1}x+\frac{1}{2}\int\frac{1}{\sqrt{u}}du=\sin^{-1}x+\frac{1}{2}\int u^{-\frac{1}{2}}du\\ &=\sin^{-1}x+\sqrt{1-x^2}+C \end{aligned}$

77. $\int x^{x/ln(x)}dx$

Interesting…
$\begin{aligned} &\int x^{\frac{x}{ln(x)}} dx\\ &( (\frac{x}{ln(x)})'=\frac{ln(x)-1}{(ln(x))^2})\\ &(y=x^{\frac{x}{ln(x)}}, ln(y)=\frac{x}{ln(x)}ln(x))\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{(ln(x))^2}ln(x)+\frac{x}{ln(x)}\frac{1}{x})\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{ln(x)}+\frac{1}{ln(x)}=1)\\ &y'=y\\ &=x^{\frac{x}{ln(x)}}+C \end{aligned}$
Alt.
$\begin{aligned} &\int x^{\frac{x}{ln(x)}} dx = \int (e^{ln x})^{\frac{x}{ln(x)}}dx=\int e^x dx\\ &= e^x+C\\ &(x^{\frac{x}{ln(x)}}=e^x) \end{aligned}$

78. $\int arcsin(sqrt(x))dx$

$\begin{aligned} &\int arcsin(\sqrt x) dx \\ &( \text{ 루트부분을 치환. 부분적분} )\\ &( u = \sqrt x, u^2=x, 2udu=dx)\\ &=2\int u sin^{-1}(u) du= (arcsin은 미분가능)\\ &=2( sin^{-1}u \frac{1}{2}u^2-\int \frac{1}{\sqrt{1-u^2}}\frac{1}{2}u^2 du )\\ &=u^2sin^{-1}u-\int \frac{u^2}{\sqrt{1-u^2}}du\\ &=u^2sin^{-1}u+\int \frac{-1+1-u^2}{\sqrt{1-u^2}}du\\ &=x\sin^{-1}\sqrt{x}-\int \frac{1}{\sqrt{1-u^2}}du+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ \end{aligned}$

$\int \sqrt{1-u^2}du, (u=sin\theta, du=cos\theta d\theta)\\ =\int cos^2\theta d\theta=\frac{1}{2}\int1+cos2\theta d\theta=\frac{1}{2}\theta+\frac{1}{2}\int cos 2\theta d\theta\\ =\frac{1}{2}\theta+\frac{1}{4} sin 2\theta =\frac{1}{2}sin^{-1}u+\frac{1}{2} sin \theta cos \theta\\ =\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}$

$\begin{aligned} &\int arcsin(\sqrt x) dx \\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\frac{1}{2}sin^{-1}\sqrt{x}+\frac{1}{2} \sqrt{x} \sqrt{1-x}\\ &=(\sin^{-1}x)(x-\frac{1}{2})+\frac{1}{2}\sqrt{x(1-x)}+C \end{aligned}$

79. $\int arctan(x) dx$

$\begin{aligned} &\int arctan(x) dx \\ &(y=arctan(x), x=tan(y), dx=sec^2ydy)\\ &=\int y sec^2y dy (부분적분)\\ &=y \tan(y)-\int tan(y) dy=x\arctan{x}+ln|cos y|+C\\ &=x\arctan{x}+ln|\frac{1}{\sqrt{1+x^2}}|+C\\ \end{aligned}$
Alt.
$\begin{aligned} &\int arctan(x) dx \\ &(\text{we know} \int \frac{1}{1+x^2} dx=arctan(x))\\ &=arctan(x) x -\int \frac{x}{1+x^2} dx (x^2을 치환)\\ &=x \arctan{x}-\frac{1}{2}ln|1+x^2|+C \end{aligned}$

80. $\int_0^5 f(x) dx$

if x<=2, f(x)=10
if x>=2, f(x)=$3x^2-2$
$\begin{aligned} &=\int_0^2 f(x)dx+ \int_2^5 f(x) dx \\ &=\int_0^2 10 dx+ \int_2^5 3x^2-2 dx \\ &=20+\bigg[x^3-2x \bigg ]_2^5=20+(115-4)\\ &=131 \end{aligned}$

Author: crazyj7@gmail.com

61. $\int sqrt(x^2 + 4x)dx$

$\begin{aligned} &\int \sqrt{x^2 + 4x}dx = \int \sqrt{x^2 + 4x+4-4}dx \\ &=\int \sqrt{(x+2)^2-4}dx (x+2=2sec\theta, dx=2sec\theta tan \theta d\theta)\\ &=\int \sqrt{(4sec^2\theta-4) } 2sec\theta tan \theta d\theta\\ &=\int 4sec\theta tan^2\theta d\theta =4 \int sec \theta (sec^2\theta -1) d\theta \\ &=4\int sec^3 \theta d\theta -4\int sec\theta d\theta \end{aligned}$

$\int sec^3 x dx = \int sec x sec^2x dx=secxtanx-\int sec x tan^2 x dx\\ =sec x tan x-\int sec x (sec^2x-1)dx\\ =sec x tan x-\int sec^3x dx +\int sec x dx\\ 2\int sec^3x dx = secx tanx+ln|sec x+tan x|\\ \int sec^3 x dx = \frac{1}{2}(sec x tan x + ln |sec x + tan x|)+C$

$=4\int sec^3 \theta d\theta -4\int sec\theta d\theta\\ =2(sec \theta tan \theta + ln |sec \theta + tan \theta|)-4ln |sec \theta + tan \theta|\\ =2sec \theta tan \theta - 2ln|sec \theta + tan \theta|+C\\ (sec \theta = \frac{x+2}{2}, R.T. angle=\theta, a=2, h=x+2, o=\sqrt{x^2+4x})\\ =2 \frac{x+2}{2}\frac{\sqrt{x^2+4}}{2}-2ln|\frac{x+2+\sqrt{x^2+4}}{2}|+C\\ =\frac{(x+2)\sqrt{x^2+4}}{2}-2ln|\frac{x+2+\sqrt{x^2+4}}{2}|+C\\ =\frac{(x+2)\sqrt{x^2+4}}{2}-2ln|x+2+\sqrt{x^2+4}|+C_2\\$

62. $\int (x^2)e^{x^3}dx$

$\begin{aligned} &\int (x^2)e^{x^3} dx \\ &(u=x^3, du=3x^2dx)\\ &=\int (x^2)e^{u} \frac{1}{3x^2} du=e^u\\ &=\frac{1}{3}e^{x^3}+C \end{aligned}$

63. $\int (x^3)e^{x^2}dx$

$\begin{aligned} &\int (x^3)e^{x^2}dx (u=x^2, du=2xdx)\\ &=\frac{1}{2}\int (x^2)e^udu=\frac{1}{2}\int ue^udu\\ &=\frac{1}{2} ( ue^u-e^u ) \\ &= \frac{1}{2}(x^2-1 )e^{x^2}+C \end{aligned}$

64. $\int tan(x)ln(cos(x))dx$

$\begin{aligned} &\int tan(x)ln(cos(x)) dx = \int \frac{sin(x)}{cos(x)}ln(cos(x)) dx \\ &(u=cos(x), du=-sin(x)dx)\\ &=-\int ln(u)\frac{1}{u} du = -(ln(u)ln(u)-\int \frac{ln(u)}{u} du)\\ &[ 2\int \frac{ln(u)}{u} du = (ln(u))^2 ]\\ &=-\frac{1}{2}(ln(\cos x))^2+C \end{aligned}\\$
Alt.
$(u=ln(cos(x), du = -\frac{sin x}{cos x}dx=-tan x dx)\\ \int tan(x)ln(cos(x)) dx = -\int u du \\ =-\frac{1}{2}(ln(\cos x))^2+C$

65. $\int 1/(x^3 - 4x^2)dx$

$\begin{aligned} &\int \frac{1}{(x^3 - 4x^2)}dx \\ &=\int \frac{1}{x^2(x-4)}dx\\ &(\frac{ax+b}{x^2}+\frac{c}{x-4}=\frac{(c+a)x^2+(b-4a)x-4b}{x^2(x-4)})\\ &(b=-\frac{1}{4}, a=b/4=-\frac{1}{16}, c=-a=\frac{1}{16})\\ &=\int \frac{-\frac{1}{16}x-\frac{1}{4}}{x^2}dx+\int \frac{\frac{1}{16}}{x-4}dx\\ &=-\frac{1}{16} (\int \frac{x+4}{x^2}dx-\int \frac{1}{x-4}dx)\\ &=-\frac{1}{16} (\int \frac{1}{x}dx+4\int x^{-2}dx-ln|x-4|)\\ &=-\frac{1}{16} (ln|x|-\frac{4}{x}-ln|x-4|)+C \end{aligned}$

66. $\int sin(x)cos(2x)dx$

$\begin{aligned} &\int sin(x)cos(2x)dx =\int sin(x) (2cos^2(x)-1)dx\\ &=2\int sin(x)cos^2(x)dx-\int sin(x) dx (u=cos(x), du=-sin(x)dx)\\ &=-2\int u^2 du+cos(x)=-\frac{2}{3}u^3+cos(x)+C\\ &=-\frac{2}{3}cos^3(x)+cos(x)+C \end{aligned}$

67. $\int 2^{ln(x)} dx$

$\begin{aligned} &\int 2^{ln(x)} dx \\ &(y=2^{ln(x)}, log_2y=\frac{ln y}{ln 2}=ln(x))\\ &(\frac{1}{yln 2}dy=\frac{1}{x}dx, \frac{dy}{dx}=\frac{yln 2}{x})\\ &=\int y \frac{x}{yln 2}dy=\frac{1}{ln 2}\int x dy\\ &=\frac{1}{ln 2}\int e^{log_2y} dy\\ &=\frac{1}{ln 2}\int y^{log_2e} dy\\ &=\frac{1}{ln 2}\int y^{\frac{1}{ln 2}} dy\\ &(\frac{1}{ln 2}+1 =\frac{1+ln 2}{ln 2} )\\ &=\frac{1}{ln 2}\frac{ln2}{(1+ln 2)} y^{(\frac{1}{ln 2}+1)}\\ &=\frac{1}{(1+ln 2)} y y^{1/ln 2}\\ &=\frac{1}{(1+ln2)} 2^{ln x}2^{(ln x)(1/ln 2)}+C\\ &=\frac{1}{(1+ln2)} 2^{ln x}x^{(ln 2)(1/ln 2)}+C\\ &=\frac{x 2^{lnx}}{1+ln2}+C \end{aligned}$
Alt.
$\begin{aligned} &\int 2^{ln(x)} dx = \int x^{ln(2)} dx\\ &=\frac{1}{1+ln2}x^{(1+ln2)}+C\\ &=\frac{x x^{ln2}}{1+ln2}+C=\frac{x 2^{lnx}}{1+ln2}+C\\ \end{aligned}$

68. $\int sqrt(1+cos(2x)) dx$

$\begin{aligned} &\int \sqrt{1+cos(2x)}dx \\ &=\int \sqrt{1+2cos^2x-1}dx=\sqrt 2 \int cos x dx\\ &=\sqrt 2 \sin x +C \end{aligned}$

69. $\int 1/(1+tan(x)) x$

$\begin{aligned} &\int \frac{1}{1+tan(x)} dx \\ &=\int \frac{1-tan(x)}{1-tan^2(x)} dx=\int \frac{1+2tan(x)-3tan(x)}{1-tan^2(x)} dx\\ &=\int tan(2x) dx + \int \frac{1-3tan(x)}{1-tan^2(x)}dx\\ &=-\frac{1}{2}ln|cos 2x|+ \int \frac{\frac{cosx-3sinx}{cosx}}{\frac{cos^2x-sin^2x}{cos^2x}}dx \\ &=-\frac{1}{2}ln|cos 2x|+ \int \frac{cos^2x-3cos x sinx}{cos^2x-sin^2x}dx \\ &=\int \frac{cos x}{cos x+sin x} dx (u=sinx, du=cosxdx)\\ &=\int \frac{1}{\sqrt{1-u^2} + u} du\\ &=\int \frac{1+tan(x)}{1+tan^2(x)+2tan(x)} dx\\ &\int \frac{}{1+tan(x)} dx \\ &tan(2x)=\frac{2tan(x)}{1-tan^2(x)} \end{aligned}$
first!..
$\int \frac{1-tan(x)}{1+tan(x)} dx =\int \frac{cos-sin}{cos+sin} dx (u=cos+sin, du=-sin+cos dx)\\ =\int \frac{1}{u} du = ln|u|=ln|cos(x)+sin(x)|+C$
$\int \frac{1}{1+tan(x)} dx =\frac{1}{2}\int \frac{1-tan(x)+1+tan(x)}{1+tan(x)} dx\\ =\frac{1}{2} \int \frac{1-tan(x)}{1+tan(x)} dx +\frac{1}{2}x\\ =\frac{1}{2}ln|cos(x)+sin(x)|+\frac{1}{2}x+C$

70. $\int_{1/e}^e sqrt(1- ln(x)^2)/x dx$

$\begin{aligned} &\int_{1/e}^e \frac{\sqrt{1- ln(x)^2}}{x} dx (u=ln(x), du=\frac{1}{x} dx)\\ &=\int_{-1}^{1} \sqrt{1-u^2} du =2\int_{0}^{1} \sqrt{1-u^2} du \\ &= \text{area half circle, radius 1} \\ &=\frac{\pi}{2}\\ & (u=sin \theta, du=cos\theta d\theta)\\ &=2\int_0^{\pi/2} cos^2\theta d\theta=\int_0^{\pi/2} 1+cos2\theta d\theta\\ &=\bigg[ \theta +\frac{1}{2}sin 2\theta \bigg]_0^{\pi/2}\\ &=\frac{\pi}{2} \end{aligned}$

Author: crazyj7@gmail.com

cmd(콘솔)창 관리자권한으로 실행하기

cmd 창을 자주 사용하는데, 가끔 관리자 권한이 필요해서 다시 창을 띄우는데 띄우기가 번거롭다.
윈도우 기본 기능으로 빨리 cmd창을 관리자 권한으로 실행할 수 없을까?

1. Ctrl+Shift+ESC 키를 누른다. (작업관리자 실행됨)
2. Alt를 계속 누른 상태로 F, N을 차례로 누른다.
   
3. cmd를 입력 후, 탭키, 스페이스키, 엔터 순서로 입력한다. (관리자 권한을 체크하여 실행)

Author: crazyj7@gmail.com