Integral100 [81-89]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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