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Question Number 111189 by john santu last updated on 02/Sep/20

$$\left(1\right)\int \frac{(x+1)dx}{x^4(x-1)}?$$$$\left(2\right)\frac{dy}{dx}+\frac{y}{x-2}=5(x-2)\sqrt{y}$$

Answered by Sarah85 last updated on 02/Sep/20

$$\frac{x+1}{x^4(x-1)}=\frac{2}{x-1}-\frac{2}{x}-\frac{2}{x^2}-\frac{2}{x^3}-\frac{1}{x^4}$$$$\int \frac{x+1}{x^4(x-1)}dx=\ln \frac{{(x-1)}^2}{x^2}+\frac{6x^2+3x+1}{3x^3}+C$$

Answered by Dwaipayan Shikari last updated on 02/Sep/20

$$\int \frac{x-1}{x^4(x-1)}+\frac{2}{x^4(x-1)}$$$$-\frac{1}{3x^3}-2\int \frac{1}{x^3}(\frac{1}{x}-\frac{1}{x-1})$$$$=-\frac{1}{3x^3}+\frac{2}{3x^3}+2\int \frac{1}{x^3(x-1)}dx$$$$=\frac{1}{3x^3}-2\int \frac{1}{x^2}(\frac{1}{x}-\frac{1}{x-1})$$$$=\frac{1}{3x^3}+\frac{1}{x^2}+2\int \frac{1}{x^2(x-1)}dx$$$$=\frac{1}{3x^3}+\frac{1}{x^2}-2\int \frac{1}{x}(\frac{1}{x}-\frac{1}{x-1})$$$$=\frac{1}{3x^3}+\frac{1}{x^2}+\frac{2}{x}-2\int \frac{1}{x}-\frac{1}{x-1}$$$$=\frac{1}{3x^3}+\frac{1}{x^2}+\frac{2}{x}-2log\left(x\right)+2log(x-1)+C$$

Answered by bemath last updated on 02/Sep/20

Answered by ajfour last updated on 02/Sep/20

$$lety=t^2,x-2=s$$$$\Rightarrow \frac{2tdt}{ds}+\frac{t^2}{s}=5st$$$$\Rightarrow ift\ne 0,\frac{dt}{ds}+\frac{t}{2s}=\frac{5s}{2}$$$$\Rightarrow I.F.=e^{\frac{1}{2}\int \frac{ds}{s}}=\sqrt{s}$$$$t\sqrt{s}=\int \frac{5s\sqrt{s}}{2}$$$$t\sqrt{s}=s^2\sqrt{s}+c$$$$\Rightarrow \sqrt{y}-{(x-2)}^2=\frac{c}{\sqrt{x-2}}$$

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