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Question Number 47851 by maxmathsup by imad last updated on 15/Nov/18

$$letf\left(x\right)=x+1+\sqrt{x}andg\left(x\right)=x+1-\sqrt{x}$$$$find\int \frac{f\left(x\right)}{g\left(x\right)}dxand\frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}.$$

Commented by maxmathsup by imad last updated on 16/Nov/18

$$letI=\int \frac{f\left(x\right)}{g\left(x\right)}dx\Rightarrow I=\int \frac{x+1+\sqrt{x}}{x+1-\sqrt{x}}dxcha7gement\sqrt{x}=tgive$$$$I=\int \frac{t^2+t+1}{t^2-t+1}\left(2t\right)dt=2\int \frac{t^3+t^2+1}{t^2-t+1}dt$$$$=2\int \frac{t(t^2-t+1)+t^2-t+t^2+1}{t^2-t+1}dt=2\int tdt+2\int \frac{2t^2-t+1}{t^2-t+1}dt$$$$=t^2+2\int \frac{2(t^2-t+1)+2t-2-t+1}{t^2-t+1}dt$$$$=t^2+4t+\int \frac{2t-1-1}{t^2-t+1}dt=t^2+4t+ln\mid t^2-t+1\mid -\int \frac{dt}{t^2-t+1}but$$$$\int \frac{dt}{t^2-t+1}=\int \frac{dt}{t^2-2\frac{t}{2}+\frac{1}{4}+\frac{3}{4}}=\int \frac{dt}{{(t-\frac{1}{2})}^2+\frac{3}{4}}$$$${=}_{t-\frac{1}{2}=\frac{\sqrt{3}}{2}u}\frac{4}{3}\int \frac{1}{1+u^2}\frac{\sqrt{3}}{2}du=\frac{2}{\sqrt{3}}arctan\left(\frac{2t-1}{\sqrt{3}}\right)\Rightarrow$$$$I=t^2+4t-\frac{2}{\sqrt{3}}arctan\left(\frac{2t-1}{\sqrt{3}}\right)+C.$$

Commented by maxmathsup by imad last updated on 16/Nov/18

$$butt=\sqrt{x}\Rightarrow I=x+4\sqrt{x}-\frac{2}{\sqrt{3}}arctan\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)+C.$$

Commented by maxmathsup by imad last updated on 18/Nov/18

$$changement\sqrt{x}=tgive\int f\left(x\right)dx=\int (t^2+1+t)\left(2t\right)dt$$$$=2\int (t^3+t^2+t)dt=\frac{1}{2}{t}^4+\frac{2}{3}{t}^3+t^2+c_1=\frac{x^2}{2}+\frac{2}{3}x\sqrt{x}+x+c_1$$$$\int g\left(x\right)dx=\int (x+1-\sqrt{x})dx{=}_{\sqrt{x}=t}\int (t^2+1-t)\left(2t\right)dt$$$$=2\int (t^3+t-t^2)dt=\frac{t^4}{4}-\frac{2}{3}{t}^3+t^2+c_2=\frac{x^2}{2}-\frac{2}{3}x\sqrt{x}+x+c_2$$$$\Rightarrow \frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}=\frac{\frac{x^2}{2}+\frac{2}{3}x\sqrt{x}+x+c_1}{\frac{x^2}{2}-\frac{2}{3}x\sqrt{x}+x+c_2}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Nov/18

$$\int f\left(x\right)dx$$$$\int x+1+\sqrt{x}dx$$$$=\frac{x^2}{2}+x+\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c_1$$$$\int g\left(x\right)dx=\frac{x^2}{2}+x-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c_2$$$$so\frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}=\frac{\frac{x^2}{2}+x+\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c_1}{\frac{x^2}{2}+x-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c_2}$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Nov/18

$$\int \frac{x+1+\sqrt{x}}{x+1-\sqrt{x}}dx$$$$\int \frac{\sqrt{x}+\frac{1}{\sqrt{x}}+1}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}dx$$$$\int \frac{\sqrt{x}+\frac{1}{\sqrt{x}}-1+2}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}dx$$$$\int dx+2\int \frac{dx}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}$$$$t^2=xdx=2tdt$$$$I_1=\int dx=x+c_1$$$$I_2=\int \frac{2tdt}{t+\frac{1}{t}-1}dt$$$$\int \frac{2t^2}{t^2+1-t}dt$$$$2\int \frac{t^2-t+1+t-1}{t^2-t+1}dt$$$$2\int dt+2\int \frac{t-1}{t^2-t+1}dt$$$$2\int dt+\int \frac{2t-1-1}{t^2-t+1}dt$$$$2\int dt+\int \frac{d(t^2-t+1)}{t^2-t+1}-\int \frac{dt}{(t^2-2.t.\frac{1}{2}+\frac{1}{4}+\frac{3}{4})}$$$$2\int dt+\int \frac{d(t^2-t+1)}{t^2-t+1}-\int \frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$2t-ln(t^2-t+1)-\frac{2}{\frac{\sqrt{3}}{2}}{tan}^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$$$$2\sqrt{x}-ln(x-\sqrt{x}+1)-\frac{4}{\sqrt{3}}{tan}^{-1}\left(\frac{\sqrt{x}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$$

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