91. $\int \frac{x}{1 + x^4} dx$

$\begin{aligned} &\int \frac{x}{1 + x^4} dx \\ &u=x^2, du=2xdx\\ &=\frac{1}{2}\int \frac{1}{1+u^2}du=\frac{1}{2}arctan(u)\\ &=\frac{1}{2}arctan(x^2)+C\\ \end{aligned}$

92. $\int e^{\sqrt x} dx$

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

93. $\int \frac{1}{csc(x)^3} dx$

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

94. $\int \frac{arcsin x}{\sqrt{1 - x^2}}dx$

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

95. $\int \sqrt{1 + sin(2x)} dx$

$\begin{aligned} &\int \sqrt{1 + sin(2x)} dx \\ &u=sin(2x) , du=2cos(2x)dx\\ &=\int \sqrt {1+u} \frac{1}{2cos(2x)}du =\frac 1 2 \int \sqrt {1+u} \frac {1}{1-2sin^2(x)}du \\ &=\frac 1 2 \int \sqrt u \frac{1}{1-2(u-1)^2}du =\frac 1 2 \int \sqrt u \frac{1}{-2u^2+4u-1}du \\ & not work \end{aligned}$

$\begin{aligned} &\int \sqrt{1 + sin(2x)} dx \\ &u=\sqrt{1 + sin(2x)}, du=\frac{2cos(2x)}{2\sqrt{1 + sin(2x)}}dx\\ &u^2-1=sin(2x)\\ &=\int u\frac{u}{cos(2x)} du = \int \frac{u^2}{\sqrt{1-(u^2-1)^2}}du\\ & t=u^2-1, dt=2udu\\ &=\frac{1}{2} \int \frac{u}{\sqrt{1-t^2}}dt=\frac{1}{2}\int \frac{\sqrt{1+t}}{\sqrt{1-t}\sqrt{1+t}}dt \\ &=\frac{1}{2}\int \frac{1}{\sqrt{1-t}}dt=\frac{1}{2}\int (1-t)^{-1/2}dt=\frac 1 2 (2)(1-t)^{1/2}\\ &=\sqrt{1-t}=\sqrt{2-u^2}=\sqrt{2-(1+sin(2x))}\\ &=\sqrt{1-sin(2x)}+C\\ &=\sqrt{cos^2x+sin^2x-2sinxcos}+C=|cosx-sinx|+C \end{aligned}$
Alt.
$1=sin^2x+cos^2x\\ \int \sqrt{1 + sin(2x)} dx =\int \sqrt {sin^2x+cos^2x+2sinxcosx} dx\\ =\int (sinx+cosx)dx=-cosx+sinx+C$

96. $\int x^{1/4} dx$

$\begin{aligned} &\int x^{1/4} dx \\ &=\frac 4 5 x^{\frac 5 4}+C \end{aligned}$

97. $\int \frac{1}{1 + e^x}dx$

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

98. $\int \sqrt{1 + e^x} dx$

$\begin{aligned} &\int \sqrt{1 + e^x} dx \\ & u=\sqrt{1+e^x}, du=\frac{e^x}{2\sqrt{1+e^x}}dx\\ &=\int u \frac{2u}{e^x}du=2\int \frac {u^2-1+1}{u^2-1}du\\ &=2(u-\int \frac{1}{1-u^2} du)=2u-2arctanh(u)+C\\ &=2\sqrt{1+e^x}-2tanh^{-1}(\sqrt{1+e^x})+C\\ &=2\sqrt{1+e^x}-2\frac{1}{2}ln |{\frac{1+\sqrt{1+e^x}}{1-\sqrt{1+e^x}}}|+C\\ &=2\sqrt{1+e^x}+ln |{\frac{1-\sqrt{1+e^x}}{1+\sqrt{1+e^x}}}|+C\\ \end{aligned}$
$arctanh{x}=\frac{1}{2}\ln |\frac{1+x}{1-x}|$

99. $\int \frac{\sqrt{tan(x)}}{sin(2x)}dx$

$\begin{aligned} &\int \frac {\sqrt{tan(x)}}{sin(2x)}dx =\int \frac {\sqrt{tan(x)}}{2sin(x)cos(x)} dx\\ &u=\sqrt {tan(x)}, du=\frac{sec^2x}{2\sqrt{tanx}}dx \\ &=\int \frac {u}{2sinxcosx}\frac{2u}{sec^2x}du \\\\ &=\int \frac{2tan(x)cos(x)}{2sinx} du \\ &=\int du= u =\sqrt{tan(x)}+C \end{aligned}$

100. $\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx$

$\begin{aligned} &\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx \\ & \text{div cos} \\ &=\int \frac{secx}{secx+tanx}dx\\ & u=secx+tanx, du=(secxtanx+sec^2x )dx\\ &=\int \frac{secx}{u} \frac{1}{secx(tanx+secx)}du\\ &=\int \frac{1}{u^2}du = -\frac{1}{u}\\ &=-\frac{1}{secx+tanx}+C=-\frac{cosx}{1+sinx}+C \\ & -\frac{cosx}{1+sinx} ]_0^{\pi/2} =0-(-1)\\ &=1 \end{aligned}$
Alt.
$\begin{aligned} &\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx \\ &=\int_0^{\pi/2} \frac{1-sin(x)}{1 - sin^2(x)} dx =\int \frac{1-sin(x)}{cos^2(x)} dx\\ &=\int sec^2x-sec(x)tan(x)dx=tan(x)-sec(x)+C\\ &=tan(x)-sec(x) ]_0^{\pi/2} not solve. \\ &=\frac{sin(x)-1}{cos(x)}=-\frac{cos^2(x)}{cos(x)(1+sin(x))}=-\frac{cos(x)}{1+sin(x)} \end{aligned}$

101. $\int \frac {sin(x)} x + ln(x)cos(x) dx$

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

Author: crazyj7@gmail.com

81. $\int \frac{sin(\frac{1}{x})}{x^3}dx$

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

82. $\int \frac{x-1}{x^4-1}dx$

$\begin{aligned} &\int \frac{x-1}{x^4-1}dx \\ &=\int \frac{x-1}{(x^2+1)(x+1)(x-1)}dx= \int \frac{1}{(x^2+1)(x+1)}dx\\ & \frac{1}{(x^2+1)(x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{x+1}\\ & numerator:1= (c+a)x^2+(b+a)x+c+b\\ & a=-c, b=-a, c+b=1, c=(1/2), b=(1/2), a=(-1/2)\\ &=\int \frac{(-1/2)x+(1/2)}{x^2+1}dx+\int \frac{(1/2)}{x+1} dx\\ &=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x+1} dx \\ & (u=x^2+1, du=2xdx)\\ & (\int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}ln|u|)\\ &=-\frac{1}{4}ln(x^2+1)+\frac{1}{2}tan^{-1}x+\frac{1}{2}ln|x+1|+C \end{aligned}$

83. $\int \sqrt{1+(x-\frac{1}{4x})^2}dx$

$\begin{aligned} &\int \sqrt{1+(x-\frac{1}{4x})^2}dx \\ &(u=x-\frac{1}{4x}, du=(1+\frac{1}{4x^2})dx, u=\frac{4x^2-1}{4x})\\ &=\int \sqrt{1+u^2}\frac{4x^2}{4x^2+1}du=\int \sqrt{1+u^2}(1-\frac{1}{1+4x^2})du\\ & not solve...\\ & \int \sqrt{ x^2+\frac{1}{16x^2}+\frac{1}{2} }=\int \sqrt{ (x+\frac{1}{4x})^2} dx=\int x+\frac{1}{4x} dx\\ &=\frac{1}{2}x^2+\frac{1}{4}ln|x|+C \end{aligned}$

84. $\int \frac{e^{tan(x)}}{1-sin(x)^2}dx$

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

85. $\int \frac{arctan(x)}{x^2}dx$

$\begin{aligned} &\int \frac{arctan(x)}{x^2}dx \\ & D(arctan(x)) = \frac{1}{1+x^2} , \int x^{-2} dx=-x^{-1}\\ &=arctan(x)(-x^{-1})-\int \frac{1}{1+x^2}(-x^{-1})dx\\ &=arctan(x)(-x^{-1})+\int \frac{1}{x(1+x^2)}dx\\ &=-\frac{arctan(x)}{x}+\int \frac{c}{x}+\frac{ax+b}{x^2+1}dx\\ &(a+c)x^2+bx+c=1, c=1, b=0, a=-1\\ &=-\frac{arctan(x)}{x}+\int \frac{1}{x}dx-\int \frac{x}{x^2+1}dx\\ &=-\frac{arctan(x)}{x}+ln|x|-\frac{1}{2}ln(x^2+1)+C\\ &=-\frac{arctan(x)}{x}-\frac{1}{2}ln(\frac{1+x^2}{x^2})+C\\ \end{aligned}$

86. $\int \frac{arctan(x)}{1+x^2}dx$

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

87. $\int ln(x)^2dx$

$\begin{aligned} &\int ln(x)^2dx \\ &=ln(x)^2(x)-\int 2ln(x)(1/x)xdx\\ &=xln(x)^2-2\int ln(x) dx = xln(x)^2-2(xln|x|-x)\\ &=x( ln(x)^2-2ln|x|+2) +C \end{aligned}$

88. $\int \frac{\sqrt{x^2+4}}{x^2}dx$

$\begin{aligned} &\int \frac{\sqrt{x^2+4}}{x^2} dx \\ &u=\sqrt{x^2+4}, du=\frac{x}{\sqrt{x^2+4}}dx\\ &=\sqrt{x^2+4}(-\frac{1}{x})-\int \frac{x}{\sqrt{x^2+4}} (-\frac{1}{x})dx\\ &=-\frac{\sqrt{x^2+4}}{x}+\int \frac{1}{x^2+4}dx\\ &=-\frac{\sqrt{x^2+4}}{x}+\frac{1}{2}arctan({\frac{x}{2}})+C\\ \end{aligned}$

??? same???

Alt. $x=2tan\theta$
$dx=2sec^2\theta d\theta , R.T. angle=\theta, a=1, o=\frac{x}{2}, h=\sqrt{1+\frac{x^2}{4}}\\ =\int \frac{2sec\theta}{4tan^2\theta} 2sec^2\theta d\theta=\int sec^3 \theta csc^2\theta cos^2\theta d\theta \\ =\int \frac{1}{cos \theta sin^2 \theta } d\theta=\int \frac{cos^2 \theta+sin^2 \theta }{cos \theta sin^2 \theta } d\theta \\ =\int cot\theta csc \theta d\theta+\int sec \theta d\theta \\ = -csc \theta + ln |sec \theta+tan\theta| \\ =- \sqrt{1+\frac{x^2}{4}}\frac{2}{x}+ln| \sqrt{1+\frac{x^2}{4}}+\frac{x}{2}| \\ =-\frac{\sqrt{x^2+4}}{x}+ln|\frac{1}{2}(x+\sqrt{4+x^2})|+C\\ =-\frac{\sqrt{x^2+4}}{x}+ln|x+\sqrt{4+x^2}|+C_2\\$

89. $\int \frac{\sqrt{x + 4}}{x} dx$

$\begin{aligned} &\int \frac{\sqrt{x + 4}}{x} dx \\ & u=\sqrt{x+4}, du=\frac{1}{2u}dx \\ &=\int \frac{u}{u^2-4} 2u du=2 \int \frac{u^2-4+4}{u^2-4} du\\ &=2( \int du +4\int \frac{1}{u^2-4} du)=2u+8\int\frac{1/4}{u-2}+\frac{-1/4}{u+2} du\\ &=2u+2ln|u-2|-2ln|u+2|\\ &=2\sqrt{x+4}+2ln|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}|+C \end{aligned}$

Author: crazyj7@gmail.com

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