Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$

Question

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?

Answer 1

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 $\dfrac{x+1}{(x^2+7x-3)^3}$