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Question Number 36547 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$\int \frac{5x^4+4x^5}{{(x^5+x+1)}^2}$$

Answered by ajfour last updated on 03/Jun/18

$$=-\int \frac{(-\frac{5}{x^6}-\frac{4}{x^5})dx}{{(1+\frac{1}{x^4}+\frac{1}{x^5})}^2}=-\int \frac{dt}{t^2}$$$$=\frac{1}{t}+c=\frac{x^5}{x^5+x+1}+c.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$bah...verygood...excellent...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$sirajfourforgottoputsquarein{D}_r$$

Commented by ajfour last updated on 03/Jun/18

$$yes,verysilly!$$

Answered by MJS last updated on 03/Jun/18

$${\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\mathrm{once}\mathrm{more}$$$$\int \frac{p\left(x\right)}{q\left(x\right)}dx=\frac{p_1\left(x\right)}{q_1\left(x\right)}+\int \frac{p_2\left(x\right)}{q_2\left(x\right)}dx$$$$q_1\left(x\right)=\gcd (q\left(x\right),q^{\prime }\left(x\right))$$$$q_2\left(x\right)=\frac{q\left(x\right)}{q_1\left(x\right)}$$$$p_1,p_2(\mathrm{polynomes},\mathrm{degree}\left(p_i\right)<\mathrm{degree}\left(q_i\right))$$$$\mathrm{we}\mathrm{get}\mathrm{like}\mathrm{this}:$$$${(\int \frac{p\left(x\right)}{q\left(x\right)}dx=\frac{p_1\left(x\right)}{q_1\left(x\right)}+\int \frac{p_2\left(x\right)}{q_2\left(x\right)}dx)}^{\prime }\Rightarrow$$$$\Rightarrow \frac{p\left(x\right)}{q\left(x\right)}={\left(\frac{p_1\left(x\right)}{q_1\left(x\right)}\right)}^{\prime }+\frac{p_2\left(x\right)}{q_2\left(x\right)}$$$$\mathrm{match}\mathrm{the}\mathrm{constants}$$$$$$$$\mathrm{in}\mathrm{this}\mathrm{case}$$$$p\left(x\right)=4x^5+5x^4$$$$q\left(x\right)={(x^5+x+1)}^2$$$$q^{\prime }\left(x\right)=2(5x^4+1)(x^5+x+1)$$$$q_1\left(x\right)=q_2\left(x\right)=x^5+x+1$$$$\frac{4x^5+5x^4}{{(x^5+x+1)}^2}={\left(\frac{c_1{x}^4+c_2{x}^3+c_3{x}^2+c_{4}x+c_5}{x^5+x+1}\right)}^{\prime }+\frac{c_6{x}^4+c_7{x}^3+c_8{x}^2+c_{9}x+c_{10}}{x^5+x+1}\Rightarrow$$$$\Rightarrow c_4=-1;c_5=-1;\mathrm{all}\mathrm{other}{c}_i=0$$$$p_1\left(x\right)=-x-1$$$$p_2\left(x\right)=0$$$$$$$$\int \frac{5x^4+4x^5}{{(x^5+x+1)}^2}=-\frac{x+1}{x^5+x+1}+C$$

Commented by MJS last updated on 03/Jun/18

$$...\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{it}\mathrm{either}...$$

Commented by ajfour last updated on 03/Jun/18

$$pleasecheck.$$

Commented by MJS last updated on 03/Jun/18

$${(-\frac{x+1}{x^5+x+1})}^{\prime }=-\frac{(x^5+x+1)-(x+1)(5x^4+1)}{{(x^5+x+1)}^2}=$$$$=-\frac{-4x^5+5x^4}{{(x^5+x+1)}^2}=\frac{5x^4+4x^5}{{(x^5+x+1)}^2}$$$$...\mathrm{so}\mathrm{this}\mathrm{is}\mathrm{true}\mathrm{too}...$$$$$$$$\frac{x^5}{x^5+x+1}-1=-\frac{x+1}{x^5+x+1}\mathrm{so}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{different}C$$

Commented by ajfour last updated on 03/Jun/18

$$o^{\prime }yes!thanks.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$\int \frac{5x^4+4x^5}{{(x^5+x+1)}^2}dx$$$$\int \frac{5x^4+4x^5}{{\left\{x^5(1+\frac{1}{x^4}+\frac{1}{x^5})\right\}}^2}$$$$\int \frac{\frac{5x^4+4x^5}{x^{10}}}{{(1+\frac{1}{x^4}+\frac{1}{x^5})}^2}dx$$$$\int \frac{\frac{5}{x^6}+\frac{4}{x^5}}{{(1+\frac{1}{x^4}+\frac{1}{x^5})}^2}$$$$t=1+\frac{1}{x^4}+\frac{1}{x^5}$$$$dt=0+\frac{-4}{x^5}+\frac{-5}{x^6}dx$$$$\int \frac{-dt}{t^2}$$$$=-1\int t^{-2}dt$$$$=\frac{1}{t}+c....ans$$$$=\frac{1}{1+\frac{1}{x^4}+\frac{1}{x^5}}+cans$$$$y=\frac{x^5}{x^5+x+1}+c$$$$\frac{dy}{dx}=\frac{(x^5+x+1)5x^4-x^5(5x^4+1)}{{(x^5+x+1)}^2}$$$$=\frac{5x^9+5x^5+5x^4-5x^9-x^5}{{(x^5+x+1)}^2}$$$$=\frac{4x^5+5x^4}{{(x^5+x+1)}^2}$$$$somyansweriscorrect...$$

Commented by ajfour last updated on 03/Jun/18

$$thanks,i^{\prime }vecorrected.$$

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