An alternative way to integrate derivatives of inverse trigonometric functions? I'm getting through my book for calculus and I tried to evaluate 
$$A = \int_{-1}^{+1} {x^2 - {2 \over {x^2 + 1}}} \space dx \tag1$$
on my own, as I forgot that $\int{1 \over x^2 + 1} = \tan^{-1}(x) + c$. I wonder, however, why didn't I get the same result when going the other way (using substitution for $x^2 + 1$).
$$A = \int_{-1}^{+1} {x^2} dx - 2\int_{-1}^{+1}{1 \over {x^2 + 1}} dx \tag2$$
Choosing $u = x^2 +1$ and $du = 2x\space dx$, I turned the above into
$$A = \Bigg[\space {x^3 \over 3} \space\Bigg]_{-1}^{+1}  - 2\int_{2}^{2}{1 \over u} {1 \over {2x}}du \tag3$$
and because $2 = 2$, $A$ becomes 
$$A = {2 \over 3}\tag4$$
If I do 
$$A = {2 \over 3} + -2\Bigg[\space \tan^{-1}(x) \space\Bigg]_{-1}^{+1} = {2 \over 3} -\pi/2 - \pi/2 = {2 \over 3} - \pi\tag5$$
The answer in my book is ${2\over3} - \pi$ too. I must have $M$essed something up, but I don't know what and where. 
Would there be some ideas?
 A: Your $u$-sub is off. If $u = x^2+1$ then $du = 2x\text{d}x$. But there is no $2x$ term in your integral, so you'll need to solve for $x$ using $u$. You'll get $x = \sqrt{u-1}$, meaning $2\text{d}x = \frac{\text{d}u}{\sqrt{u-1}}$ and $$\int_{-1}^1 \frac{2}{x^2 + 1}\space \text{d}x = \int \frac{1}{u\sqrt{u-1}}\text{d}u$$ This integral in terms of $u$ is also not so easy to integrate. 

Instead, I would begin with $\int_{-1}^1 \frac{2}{x^2 + 1}\space \text{d}x$ and use trig-sub. Imagine a right triangle with angle $\theta$, opposite leg length $x$ and adjacent leg length of $1$. Then the hypotenuse of this triangle will have length $\sqrt{x^2+1}$, meaning:
$$\sec^2(\theta) =x^2+1 \\ \tan(\theta) = x \\ \sec^2(\theta)d\theta = dx$$ and plugging all this in yields $$ \int_{-1}^1 \frac{2}{x^2 + 1}\space \text{d}x = 2\int_{/}^/ \frac{1}{\sec^2(\theta)}\cdot \sec^2(\theta)d\theta  =2\int_{/}^/ d\theta  =2\Bigg[\theta +C\Bigg]_{/}^/$$ from the equation $\tan(\theta) = x$ we then know $\theta = \arctan(x)$, so $$2\Bigg[\theta +C\Bigg]_{/}^/ = 2\Bigg[\arctan(x)+C\Bigg]_{-1}^1$$ yielding the expected result.
