Evaluate $$\int\frac{x\sin x}{\left(x+\cos x\right)^2}dx$$
I could easily do $$\int\frac{x\cos x}{(x+\cos x)^2}dx$$ by adding and subtracting $\cos^2x$ in the numerator
$$\int\frac{x\cos x+\cos^2x-(1-\sin x)(1+\sin x)}{(x+\cos x)^2}dx=\int\frac{\cos x}{x+\cos x}dx-\int\frac{1-\sin x}{(x+\cos x)^2}(1+\sin x)dx$$$$=\int\frac{\cos x}{x+\cos x}dx+\frac{1+\sin x}{x+\cos x}-\int\frac{\cos x}{x+\cos x}dx=\frac{1+\sin x}{x+\cos x}+c$$
But when I do a similar procedure in given integral
$$\int\frac{x\sin x}{(x+\cos x)^2}dx=\int\frac{x\sin x+\sin x\cos x-\sin x\cos x}{(x+\cos x)^2}dx$$
$$=\int\frac{\sin x}{x+\cos x}dx-\int\frac{(1-(1-\sin x))\cos x}{(x+\cos x)^2}dx$$$$=\int\frac{\sin x}{x+\cos x}dx-\int\frac{\cos x}{(x+\cos x)^2}dx-\frac{\cos x}{x+\cos x}-\int\frac{\sin x}{x+\cos x}dx$$
$$=-\frac{\cos x}{x+\cos x}-\int\frac{\cos x}{(x+\cos x)^2}dx$$
This is what I end up getting