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Question Number 59182 by maxmathsup by imad last updated on 05/May/19

$$letf\left(x\right)=arctan(1+ix)$$$$determineRe\left(f\left(x\right)\right)andIm\left(f\left(x\right)\right)dx$$

Commented by maxmathsup by imad last updated on 10/Jun/19

$$wehave{f}^{{}^{\prime }}\left(x\right)=\frac{i}{1+{(1+ix)}^2}=\frac{i}{1+1+2ix-x^2}=\frac{i}{-x^2+2ix+2}$$$$=\frac{-i}{x^2-2ix-2}=\frac{-i}{x^2-2-2ix}=\frac{-i(x^2-2+2ix)}{{(x^2-2)}^2+4x^2}=\frac{2x}{{(x^2-2)}^2+4x^2}-i\frac{x^2}{{(x^2-2)}^2+4x^2}$$$$\Rightarrow f\left(x\right)=\int \frac{2xdx}{{(x^2-2)}^2+4x^2}+c_0-i(\int \frac{x^{2}dx}{{(x^2-2)}^2+4x^2}+c_1)\Rightarrow$$$$Re\left(f\left(x\right)\right)=\int \frac{2xdx}{{(x^2-2)}^2+4x^2}+c_{0}andIm\left(f\left(x\right)\right)=\int \frac{x^{2}dx}{{(x^2-2)}^2+4x^2}+c_1$$$$c_{0}and{c}_{1}reals$$$$\int \frac{2xdx}{{(x^2-2)}^2+4x^2}=\int \frac{2xdx}{x^4-4x^2+4+4x^2}=\int \frac{2xdx}{x^4+4}$$$$x^4+4={\left(x^2\right)}^2+2^2={(x^2+2)}^2-4x^2=(x^2+2-2x)(x^2+2+2x)\Rightarrow$$$$F\left(x\right)=\frac{2x}{x^4+4}=\frac{2x}{(x^2-2x+2)(x^2+2x+2)}=\frac{ax+b}{x^2-2x+2}+\frac{cx+d}{x^2+2x+2}\Rightarrow$$$$\int F\left(x\right)dx[=\int \frac{ax+b}{x^2-2x+2}dx+\int \frac{cx+d}{x^2+2x+2}dx=....becontinued....$$

Commented by mathmax by abdo last updated on 22/Jun/19

$$\int \frac{2xdx}{{(x^2-2)}^2+4x^2}{=}_{x^2=t}\int \frac{dt}{{(t-2)}^2+4t}=\int \frac{dt}{t^2-4t+4+4t}=\int \frac{dt}{4+t^2}$$$${=}_{t=2u}\int \frac{2du}{4+4u^2}=\frac{1}{2}\int \frac{du}{1+u^2}=\frac{1}{2}arctan\left(u\right)+c=\frac{1}{2}arctan\left(\frac{x^2}{2}\right)+c\Rightarrow$$$$Re\left(f\left(x\right)\right)=\frac{1}{2}arctan\left(\frac{x^2}{2}\right)+c_0$$$$$$

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