Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$ Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$
Is my solution correct:
Answer given in the book : $y(x+1)=c_1+c_2\log(x+1)+\log3(x+1)$


For my later reference: link wolfram alpha

Robert pointed out your mistake..
Here is another approach
$$(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$$
divide by $x+1$
$$(x+1)^2y''+3(x+1)y'+y=6\frac {\log(x+1)}{x+1}$$
on the left there is a derivative
$$((x+1)^2y')'+((x+1)y)'=6\frac {\log(x+1)}{x+1}$$
Integrate
$$(x+1)^2y'+(x+1)y=6\int \frac {\log(x+1)}{x+1} dx$$
$$(x+1)^2y'+(x+1)y=3\log^2(x+1)+K_1$$
Divide again by $x+1$
$$(x+1)y'+y=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$
$$((x+1)y)'=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$
Integrate again
$$(x+1)y=3\int\frac {\log^2(x+1)}{x+1}+K_1 \int  \frac {dx} {x+1}$$
$$(x+1)y=\log^3(x+1)+K_1 \ln |{x+1}|+K_2$$
$$ \boxed {y=\frac {\log^3(x+1)}{x+1}+K_1 \frac {\ln |{x+1}|}{x+1}+\frac {K_2}{x+1}}$$
