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Question Number 77603 by kaivan.ahmadi last updated on 08/Jan/20

$$\int \frac{10x^2-8x+1}{x^4-x^3-x+1}dx$$$$$$

Commented by mathmax by abdo last updated on 08/Jan/20

$$letI=\int \frac{10x^2-8x+1}{x^4-x^3-x+1}dx\Rightarrow I=\int F\left(x\right)dxwehave{x}^4-x^3-x+1=$$$$x^3(x-1)-(x-1)=(x-1)(x^3-1)={(x-1)}^2(x^2+x+1)\Rightarrow$$$$F\left(x\right)=\frac{10x^2-8x+1}{{(x-1)}^2(x^2+x+1)}=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{cx+d}{x^2+x+1}$$$$b={(x-1)}^{2}F\left(x\right){\mid }_{x=1}=\frac{3}{3}=1$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{{(x-1)}^2}+\frac{-ax+d}{x^2+x+1}$$$$F\left(0\right)=1=-a+1+d\Rightarrow d=a\Rightarrow F\left(x\right)=\frac{a}{x-1}+\frac{1}{{(x-1)}^2}+\frac{-ax+a}{x^2+x+1}$$$$F(-1)=\frac{19}{4}=-\frac{a}{2}+\frac{1}{4}+\frac{2a}{1}\Rightarrow 19=-2a+1+8a\Rightarrow 18=6a\Rightarrow a=3$$$$\Rightarrow F\left(x\right)=\frac{3}{x-1}+\frac{1}{{(x-1)}^2}+\frac{-3x+3}{x^2+x+1}\Rightarrow$$$$I=3\int \frac{dx}{x-1}+\int \frac{dx}{{(x-1)}^2}-\frac{3}{2}\int \frac{2x-2}{x^2+x+1}dx$$$$=3ln\mid x-1\mid -\frac{1}{x-1}-\frac{3}{2}\int \frac{2x+1-3}{x^2+x+1}dx$$$$=3ln\mid x-1\mid -\frac{1}{x-1}-\frac{3}{2}ln(x^2+x+1)+\frac{9}{2}\int \frac{dx}{x^2+x+1}wehave$$$$\int \frac{dx}{x^2+x+1}=\int \frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}{=}_{x+\frac{1}{2}=\frac{\sqrt{3}}{2}u}\frac{4}{3}\int \frac{1}{u^2+1}\times \frac{\sqrt{3}}{2}du$$$$=\frac{2}{\sqrt{3}}arctan\left(\frac{2x+1}{\sqrt{3}}\right)\Rightarrow$$$$I=3ln\mid x-1\mid -\frac{1}{x-1}+3\sqrt{3}arctan\left(\frac{2x+1}{\sqrt{3}}\right)+C.$$

Answered by Kunal12588 last updated on 08/Jan/20

$$x^4-x^3-x+1=x(x^3-1)-(x^3-1)$$$$=(x-1)(x^3-1)$$$$={(x-1)}^2(x^2+x+1)$$$$\frac{10x^2-8x+1}{{(x-1)}^2(x^2+x+1)}=\frac{a}{(x-1)}+\frac{b}{{(x-1)}^2}+\frac{cx+d}{(x^2+x+1)}$$$$\Rightarrow 10x^2-8x+1=a(x^4-x^3-x+1)+b(x^3-1)$$$$+(cx+d){(x-1)}^3$$$$10x^2-8x+1=\{\begin{array}{l}{ax}^4+(-a+b)x^3-ax+(a-b)\\ +(cx+d)(x^3-3x^2+3x-1)\end{array}$$$$10x^2-8x+1=\{\begin{array}{l}{ax}^4+(-a+b)x^3-ax+(a-b)\\ +{cx}^4+(-3c+d)x^3+(3c-3d)x^2\\ +(-c+3d)x-d\end{array}$$$$10x^2-8x+1=\{\begin{array}{l}(a+c)x^4+(-a+b-3c+d)x^3\\ +(3c-3d)x^2+(-a-c+3d)x\\ +(a-b-d)\end{array}$$$$1)a+c=0$$$$2)-a+b-3c+d=0$$$$3)3c-3d=10$$$$4)-a-c+3d=-8$$$$5)a-b-d=1$$$$1\Rightarrow c=-a$$$$\Rightarrow \{\begin{array}{l}b+d=-2a\\ -3a-3d=10\Rightarrow d=-\frac{3a+10}{3}\\ d=\frac{-8}{3}\end{array}$$$$\Rightarrow \{\begin{array}{l}b=-2a+\frac{8}{3}\\ 3a+10=8\Rightarrow a=6\Rightarrow c=-6\end{array}$$$$\Rightarrow b=-12+\frac{8}{3}=\frac{-36+8}{3}=\frac{-28}{3}$$$$\frac{10x^2-8x+1}{{(x-1)}^2(x^2+x+1)}=\frac{6}{(x-1)}-\frac{28}{3{(x-1)}^2}-\frac{18x+8}{3(x^2+x+1)}$$$$\int \frac{10x^2-8x+1}{x^4-x^3-x+1}dx$$$$=6\int \frac{dx}{x-1}-\frac{28}{3}\int \frac{dx}{{(x-1)}^2}-\frac{1}{3}\int \frac{18x+9-1}{x^2+x+1}dx$$$$=6log\mid x-1\mid +\frac{28}{3(x-1)}-\frac{9}{3}log\mid x^2+x+1\mid +\frac{1}{3}\int \frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$$$$=6log\mid x-1\mid +\frac{28}{3(x-1)}-3log\mid x^2+x+1\mid +\frac{1}{3}\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$=6log\mid x-1\mid +\frac{28}{3(x-1)}-3log\mid x^2+x+1\mid +\frac{1}{3}\times \frac{2}{\sqrt{3}}{tan}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$$$$=6log\mid x-1\mid +\frac{28}{3(x-1)}-3log\mid x^2+x+1\mid +\frac{2}{3\sqrt{3}}{tan}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$$

Commented by Kunal12588 last updated on 08/Jan/20

$$nexttimetryyourselffirst!$$

Commented by key of knowledge last updated on 08/Jan/20

$$\log \mathrm{is}\mathrm{not},\ln \mathrm{is}\mathrm{true}(\int \frac{\mathrm{du}}{\mathrm{u}}=\mathrm{lnx})$$

Commented by kaivan.ahmadi last updated on 08/Jan/20

$$thankusir$$

Commented by MJS last updated on 08/Jan/20

$$\mathrm{some}\mathrm{write}\log \mathrm{for}\ln ,\mathrm{and}{\log }_{10}$$$$\mathrm{others}\mathrm{write}\log \mathrm{for}{\log }_{10}\mathrm{and}\ln$$$$\mathrm{in}\mathrm{case}\mathrm{of}\mathrm{integrals}\mathrm{there}\mathrm{had}\mathrm{been}\mathrm{no}\mathrm{case}$$$$\mathrm{yet}\mathrm{where}\log \mathrm{meant}{\log }_{10}$$

Commented by mathmax by abdo last updated on 08/Jan/20

$$bynotationLogmeanslnandlogmeans{log}_{10}...$$

Commented by Kunal12588 last updated on 09/Jan/20

$$yeah,thereismuchofconfusion$$$$butIamjustfollowingwhatIlearnedin$$$$myschoolandNCERTbook.$$$$loga\to {log}_{e}aandforreferringto{log}_{10}a$$$$wewriteitlikethat{log}_{10}a.$$$$nothingmentionedaboutln.$$

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