I need help with the indefinite integral \begin{align} & \int\frac{dx}{(x^2+2x+5)^2} \\[10pt] = {} & \int\frac{dx}{((x+1)^2+4)^2} = \int\frac{du}{(u^2+4)^2} & & x+1=u,\quad du=dx \\[10pt] = {} & \frac{1}{16} \int\frac{du}{(\frac{u^2}{4}+1)^2} \\[10pt] = {} & \frac 1 8 \int \frac{ds}{(s^2+1)^2} = \text{?} & & \frac u 2 = s, \quad 2\,ds=du \end{align}
or maybe there is an easier way ?
Any ideas ? thanks !
it is $$x^2+2t+5=(x+1)^2+4$$ we Substitute $$x+1=t$$ then we have $$dx=dt$$ and $$\int\frac{1}{(t^2+4)^2}dt$$ and then we Substitute $$t=2\tan(s)$$ with $$dt=2\sec^2(s)ds$$ and we get $$(t^2+4)^2=16\sec^4(s)$$ and our integral is $$2\int \frac{\cos^2(s)}{16}ds$$ and in the last step note that $$\cos^2(s)=\frac{1}{2}\cos(2s)+\frac{1}{2}$$
If you write $s = \tan(t)$ then you obtain $\sec^2(t)\,\mathrm{d}t = \mathrm{d} s$ and $$ \int \frac{1}{(1+s^2)^2}\,\mathrm{d}{s}=\int \cos^2(t)\,\mathrm{d}{t} $$ you can then solve the second integral by using the double-angle formula for cosine.
There is a reduction process. Apply integration by parts.
$$\int \frac{1}{1+x^2}\,dx=\frac{x}{1+x^2}-\int x \,d\frac{1}{1+x^2}$$ $$=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}\,dx-2\int \frac{1}{(1+x^2)^2}\,dx$$ And rearranging, $$2\int \frac{1}{(1+x^2)^2}\,dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}\,dx$$
