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Question Number 36677 by mondodotto@gmail.com last updated on 04/Jun/18

Commented by prof Abdo imad last updated on 04/Jun/18

$$3)letdecomposeF\left(x\right)=\frac{2x^3-4x-8}{(x^2-x)(x^2+4)}$$$$F\left(x\right)=\frac{2x^3-4x-8}{x(x-1)(x^2+4)}=\frac{a}{x}+\frac{b}{x-1}+\frac{cx+d}{x^2+4}$$$$a={lim}_{x\to 0}xF\left(x\right)=\frac{-8}{-4}=2$$$$b={lim}_{x\to 1}(x-1)F\left(x\right)=\frac{-10}{5}=-2\Rightarrow$$$$F\left(x\right)=\frac{2}{x}-\frac{2}{x-1}+\frac{cx+d}{x^2+4}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=2=c\Rightarrow$$$$F\left(x\right)=\frac{2}{x}-\frac{2}{x-1}+\frac{2x+d}{x^2+4}$$$$F\left(2\right)=0=1-2+\frac{4+d}{8}=-1+\frac{4+d}{8}=\frac{-4+d}{8}$$$$\Rightarrow d=4\Rightarrow$$$$F\left(x\right)=\frac{2}{x}-\frac{2}{x-1}+\frac{2x+4}{x^2+4}\Rightarrow$$$$\int F\left(x\right)dx=2ln\mid x\mid -2ln\mid x-1\mid +\int \frac{2x}{x^2+4}$$$$4\int \frac{dx}{x^2+4}=2ln\mid x\mid -2ln\mid x-1\mid +ln(x^2+4)$$$$+4\int \frac{dx}{x^2+4}but\int \frac{dx}{x^2+4}{=}_{x=2t}\int \frac{1}{4(1+t^2)}\frac{dt}{2}$$$$=\frac{1}{8}arctan\left(\frac{x}{2}\right)\Rightarrow$$$$\int F\left(x\right)dx=2ln\mid x\mid -2ln\mid x-1\mid +ln(x^2+4)$$$$+\frac{1}{2}arctan\left(\frac{x}{2}\right).$$

Commented by abdo mathsup last updated on 04/Jun/18

$$1)letI=\int \frac{dx}{1+e^x}changement{e}^x=tgive$$$$I=\int \frac{1}{1+t}\frac{dt}{t}=\int (\frac{1}{t}-\frac{1}{1+t})dt$$$$=ln\mid \frac{t}{1+t}\mid +c=ln\mid \frac{e^x}{e^x+1}\mid +c$$$$=x-ln(1+e^x)+c.$$$$$$

Commented by Cheyboy last updated on 04/Jun/18

$$1.\int \frac{1}{1+{\mathit{e}}^{\mathit{x}}}\mathit{d}\mathit{x}$$$$\int \frac{1+{\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{\mathit{x}}}{1+{\mathit{e}}^{\mathit{x}}}$$$$\int (\frac{1+{\mathit{e}}^{\mathit{x}}}{1+{\mathit{e}}^{\mathit{x}}}-\frac{{\mathit{e}}^{\mathit{x}}}{1+{\mathit{e}}^{\mathit{x}}})\mathit{d}\mathit{x}$$$$\int (1-\frac{{\mathit{e}}^{\mathit{x}}}{1+{\mathit{e}}^{\mathit{x}}})$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{u}=1+{\mathit{e}}^{\mathit{x}}$$$$\mathit{d}\mathit{u}={\mathit{e}}^{\mathit{x}}\mathit{d}\mathit{x}$$$$\mathit{x}-\int \frac{1}{\mathit{u}}\mathit{d}\mathit{u}$$$$\mathit{x}-\mathit{l}\mathit{n}\mid u\mid$$$$x-\mathit{l}\mathit{n}\mid 1+{\mathit{e}}^{\mathit{x}}\mid +\mathit{C}$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

$$\int \frac{5x^2+20x+6}{x(x+1)(x+1)}dx$$$$=\frac{5x^2+20x+6}{x(x+1)(x+1)}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$5x^2+20x+6=a(x^2+2x+1)+b(x^2+x)+cx$$$$do=x^2(a+b)+x(2a+b+c)+a$$$$5x^2+20x+6=x^2(a+b)+x(2a+b+c)+a$$$$a=6$$$$a+b=5$$$$b=5-6=-1$$$$12-1+c=20$$$$c=9$$$$\int \frac{6}{x}dx+\int \frac{-1}{x+1}dx+\int \frac{9}{{(x+1)}^2}$$$$6lnx-ln(x+1)-\frac{9}{x+1}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

$$1)\int \frac{e^{-x}}{e^{-x}+1}dx$$$$=t=e^{-x}+1$$$$dt=-e^{-x}dx$$$$\int \frac{-dt}{t}$$$$-lnt+c$$$$-ln(e^{-x}+1)+c$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jun/18

$$\int \frac{2x^3-4x-8}{(x^2-x)(x^2+4)}dx$$$$\frac{2x^3-4x-8}{x(x-1)(x^2+4)}=\frac{a}{x}+\frac{b}{x-1}+\frac{cx+d}{x^2+4}$$$$2x^3-4x-8=a(x^3+4x-x^2-4)+b(x^3+4x)$$$$+(cx+d)(x^2-x)$$$$LHS=a(x^3-x^2+4x-4)+b(x^3+4x)+({cx}^3-{cx}^2$$$$+{dx}^2-dx)$$$$LHS=x^3(a+b+c)+x^2(-a-c+d)+x(4a+4b-d)$$$$+(-4a)$$$$-4a=-8$$$$a=2$$$$-2-c+d=0d=2+c$$$$8+4b-d=-4$$$$2+b+c=2$$$$b+c=0$$$$b=-c$$$$8+4b-d=-4$$$$8-4c-2-c=-4$$$$-5c=-4-8+2$$$$c=2$$$$b=-2$$$$d=2+2=4$$$$a=2b=-2c=2d=4$$$$\int \frac{2}{x}dx+\int \frac{-2}{x-1}dx+\int \frac{2x+4}{x^2+4}$$$$lnx-2ln(x-1)+\int \frac{2x}{x^2+4}dx+4\int \frac{dx}{x^2+4}$$$$lnx-2ln(x-1)+ln(x^2+4)+\frac{4}{\frac{}{}}\times \frac{1}{2}{tan}^{-1}\left(\frac{x}{2}\right)$$

Answered by MJS last updated on 04/Jun/18

$$6.$$$$\int \frac{{\sec }^{2}x}{\tan x(\tan x+1)}dx=$$$$[t=\tan x\to dx=\frac{dt}{{\sec }^{2}x}]$$$$=\int \frac{dt}{t(t+1)}=\int \frac{\mathcal{A}}{t}dt+\int \frac{\mathcal{B}}{t+1}dt=$$$$=\int \frac{dt}{t}-\int \frac{dt}{t+1}=\ln t-\ln (t+1)=$$$$=\ln \mid \tan x\mid -\ln \mid \tan x+1\mid +C$$$$$$$$7.$$$$\int \frac{3\cos x}{{\sin }^{2}x+\sin x-2}dx=$$$$[t=\sin x\to dx=\frac{dt}{\cos x}]$$$$=\int \frac{3}{t^2+t-2}dt=\int \frac{3}{(t-1)(t+2)}dt=$$$$=\int \frac{dt}{t-1}-\int \frac{dt}{t+2}=\ln (t-1)-\ln (t+2)=$$$$[\mid \sin x-1\mid =1-\sin x]$$$$=\ln (1-\sin x)-\ln (2+\sin x)+C$$

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