Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How can i solve something like that? 
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
How should I start? Should I try rewrite it in partial fractions?

HINT:
As $x^2+7x-3=\dfrac{(2x+7)^2-61}4$ start with $2x+7=\sqrt{61}\sec y$
Alternatively,
as the highest power of $x$ in $(x^2+7x-3)^2$ is $4,$
differentiate  $\dfrac{ax^3+bx^2+cx+d}{(x^2+7x-3)^2}$  wrt $x$
and compare with $\dfrac{x+1}{(x^2+7x-3)^3}$
