expressing $\int tan^{-1}tdt$ as a power series I have a question which asks me to state $\int_0^x\tan^{-1}(t)dt$ as a power series in $x$ and then use that result to show that $\frac{\pi}{4}-\log\sqrt{2}= 1-1/2-1/3+1/4+1/5--++\cdots$
For the first part:
$$\tan^{-1}t= \int\sum_{n=0}^\infty(-1)^nx^{2n}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\cdots$$6
then integrating from $0$ to $x$ we get 
$$\int_0^x\tan^{-1}(t)dt= \frac{x^2}{2}-\frac{x^4}{3\cdot4}+\frac{x^5}{5\cdot6}-\frac{x^8}{7\cdot8}\cdots=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{n(n-1)}$$
But how can I use this to show the second part?

You can use the integration-by-parts formula to get the second:
\begin{eqnarray*}
\int_0^x\tan^{-1}tdt&=&t\tan^{-1}t\big|_0^x-\int_0^x\frac{t}{t^2+1}dt\\
&=&t\tan^{-1}t\big|_0^x-\frac{1}{2}\ln(t^2+1)\big|_0^x\\
&=&x\tan^{-1}x-\frac{1}{2}\ln(x^2+1).
\end{eqnarray*}
So
$$ \int_0^1\tan^{-1}tdt=\frac{\pi}{4}-\frac{1}{2}\ln 2=\cdots. $$
