Each of these ODEs is linear. Isolate [tex]y'[/tex] and find the corresponding integrating factors.
[tex]\implies y'+\dfrac{1+2x^2}xy=x^2e^{-x^2}[/tex]
The integrating factor in this case is [tex]\mu[/tex] where
[tex]\ln\mu=\displaystyle\int\frac{1+2x^2}x\,\mathrm dx=\int\left(\frac1x+2x\right)\,\mathrm dx[/tex]
[tex]\implies\ln\mu=\ln x+x^2\implies\mu=xe^{x^2}[/tex]
Multiplying both sides of the ODE by [tex]\mu[/tex] gives
[tex]xe^{x^2}y'+(1+2x^2)e^{x^2}y=x^3[/tex]
Notice that [tex](xe^{x^2})'=(1+2x^2)e^{x^2}[/tex]; condensing the left side gives
[tex](xe^{x^2}y)'=x^3[/tex]
[tex]\implies xe^{x^2}y=\dfrac{x^4}4+C[/tex]
[tex]\implies\boxed{y=\dfrac{x^3}{4e^{x^2}}+Ce^{-x^2}}[/tex]
[tex]\implies y'+\dfrac2xy=\dfrac2{x^2}+1[/tex]
[tex]\ln\mu=\displaystyle\int\frac2x\,\mathrm dx=2\ln x\implies\mu=x^2[/tex]
[tex]x^2y'+2xy=2+x^2\implies(x^2y)'=2+x^2\implies x^2y=2x+\dfrac{x^3}3+C[/tex]
[tex]\implies\boxed{y=\dfrac2x+\dfrac x3+\dfrac C{x^2}}[/tex]
[tex]\ln\mu=\displaystyle\int\frac4{x-1}\,\mathrm dx=4\ln(x-1)\implies\mu=(x-1)^4[/tex]
[tex](x-1)^4y'+4(x-1)^3y=\dfrac1{x-1}+\sin x[/tex]
[tex]\implies((x-1)^4y)'=\dfrac1{x-1}+\sin x[/tex]
[tex]\implies(x-1)^4y=\ln|x-1|-\cos x+C[/tex]
[tex]\implies\boxed{y=\dfrac{\ln|x-1|-\cos x+C}{(x-1)^4}}[/tex]
