Integrate $\int \frac{(x+1)}{(x^2+2)^2}dx$ This is the question I want to ask
Integrate $$\int \frac{(x+1)}{(x^2+2)^2}\,dx.$$
 I tried it using algebric manipulation
Integrate $x/(x^2+2)^2+1/(x^2+2)^2$.
Then the latter part would not be solved.

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{1}{(x^2+2)^2}dx$$
Let $x=\sqrt{2}\tan \theta\;,$ Then $dx = \sqrt{2}\sec^2 \theta d\theta$
So Integral $$\displaystyle I = \int\frac{\sqrt{2}\sec^2 \theta }{4\sec^4 \theta }d\theta = \frac{1}{2\sqrt{2}}\int \cos ^2\theta d\theta = \frac{1}{4\sqrt{2}}\int (1+\cos 2 \theta )d\theta$$
So $$\displaystyle I = \frac{1}{4\sqrt{2}}\left(\theta +\frac{\sin 2\theta }{2}\right)+\mathcal{C} = \frac{1}{4\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{4\sqrt{2}}\cdot \frac{\sqrt{2}x}{x^2+2}+\mathcal{C}$$
So $$\displaystyle I = \int \frac{1}{(x^2+2)^2}dx = \frac{1}{4\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{4}\cdot \frac{x}{x^2+2}+\mathcal{C}$$
