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Question Number 43322 by Raj Singh last updated on 09/Sep/18

Commented by maxmathsup by imad last updated on 11/Sep/18

$$letA=\int \frac{dx}{{(x+1)}^2(x^2+1)}letdecomposeF\left(x\right)=\frac{1}{{(x+1)}^2(x^2+1)}$$$$F\left(x\right)=\frac{a}{x+1}+\frac{b}{{(x+1)}^2}+\frac{cx+d}{x^2+1}$$$$b={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=\frac{1}{2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-\frac{1}{2}\Rightarrow F\left(x\right)=\frac{a}{x+1}+\frac{1}{2{(x+1)}^2}+\frac{-\frac{1}{2}x+d}{x^2+1}$$$$F\left(o\right)=1=a+\frac{1}{2}+d\Rightarrow a+d=\frac{1}{2}$$$$F\left(2\right)=\frac{1}{45}=\frac{a}{3}+\frac{1}{18}+\frac{d-1}{5}\Rightarrow 1=15a+\frac{45}{18}+9d-9\Rightarrow$$$$1=15a+\frac{5}{2}+9d-9\Rightarrow 15a+9d=10-\frac{5}{2}=\frac{15}{2}\Rightarrow$$$$15a+9(\frac{1}{2}-a)=\frac{15}{2}\Rightarrow 6a=\frac{15}{2}-\frac{9}{2}=3\Rightarrow a=\frac{1}{2}\Rightarrow d=0\Rightarrow$$$$F\left(x\right)=\frac{1}{2(x+1)}+\frac{1}{2{(x+1)}^2}-\frac{x}{2(1+x^2)}\Rightarrow$$$$A=\int F\left(x\right)dx=\frac{1}{2}ln\mid x+1\mid -\frac{1}{2(x+1)}-\frac{1}{4}ln(1+x^2)+c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18

$$2)\int \frac{dx}{{(x+1)}^2(x^2+1)}$$$$\int \frac{a}{(x+1)}dx+\int \frac{b}{{(x+1)}^2}dx+\int \frac{cx+d}{x^2+1}dx$$$$nowcalculatingabcandd$$$$\frac{1}{{(x+1)}^2(x^2+1)}=\frac{a}{x+1}+\frac{b}{{(x+1)}^2}+\frac{cx+d}{x^2+1}$$$$1=a(x+1)(x^2+1)+b(x^2+1)+(cx+d){(x+1)}^2$$$$1=a(x^3+x^2+x+1)+b(x^2+1)+(cx+d)(x^2+2x+1)$$$$1=a(x^3+x^2+x+1)+b(x^2+1)+(\mathrm{c}{x}^3+2{cx}^2+cx+{dx}^2+2dx+d)$$$$1=x^3(a+c)+x^2(a+b+2c+d)+x(a+c+2d)+(a+b+d)$$$$a+c==0$$$$a+b+2c+d=0$$$$$$$$a+c+2d=0$$$$a+b+d=1$$$$a+c+d=0$$$$0+d=0d=0$$$$a+b=1$$$$a+b+2c+d=0$$$$1+2c+0=0$$$$c=\frac{-1}{2}a+c=0a=-c$$$$a=\frac{1}{2}a=\frac{1}{2}a+b=1b=\frac{1}{2}c=\frac{-1}{2}d=0$$$$\int \frac{a}{x+1}dx+\int \frac{b}{{(x+1)}^2}+\int \frac{cx}{x^2+1}dx$$$$\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{dx}{{(x+1)}^2}+\frac{-1}{2}\int \frac{dx}{x^2+1}$$$$\frac{1}{2}ln(x+1)+\frac{1}{2}\times \frac{{(x+1)}^{-1}}{-1}-\frac{1}{2}{tan}^{-1}\left(x\right)+c$$$$\frac{1}{2}ln(x+1)-\frac{1}{2(x+1)}-\frac{1}{2}{tan}^{-1}\left(x\right)+c$$$$$$

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