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Prove It The first two have been taken care of.
for $\displaystyle \int{\frac{1}{1 + 4x^2}\,dx}$ you do not use a $\displaystyle u$ substitution. You use a trigonometric substitution.
$\displaystyle \int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}$
$\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}$.
Now use the substitution $\displaystyle x = \frac{1}{2}\tan{\theta}$.
Note that $\displaystyle \frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}$.
So $\displaystyle dx = \frac{1}{2}\sec^2{\theta}\,d\theta$.
Therefore
$\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{ 2}\sec^2{\theta}\,d\theta}$
$\displaystyle = \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\ theta}\,d\theta}$
$\displaystyle = \frac{1}{2}\int{1\,d\theta}$
$\displaystyle = \frac{1}{2}\theta + C$.
Now remembering that $\displaystyle x = \frac{1}{2}\tan{\theta}$
$\displaystyle 2x = \tan{\theta}$
$\displaystyle \theta = \arctan{2x}$.
So finally our answer is $\displaystyle \frac{1}{2}\arctan{2x} + C$.