Question

Integrate for y :$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x$.

Answer 1

Given,

$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x$

$dy=\frac{2x^2+x}{x^3+x^2+x+1}dx$

integrating on both sides, we get,

$\int dy=\int \frac{2x^2+x}{x^3+x^2+x+1}dx$

$y=\int \frac{2x^2+x}{x^3+x^2+x+1}dx$

$y=\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$

$\frac{2x^2+x}{(x+1)(x^2+1)}dx=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$

$2x^2+x=A(x^2+1)+(Bx+C)=(x+1)$

by putting $x=-1$ we get, $A=\frac{1}{2}$

by putting $x=1$ we get, $B=\frac{3}{2}$

by putting $x=0$ we get, $C=-\frac{1}{2}$

$\therefore y=\frac{1}{2(x+1)}+\frac{3x-1}{2(x^2+1)}$

$\Rightarrow y=\frac{1}{2}\log (x+1)+\frac{3}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+c$