Can somebody help me writing $\int \frac{x-\tan^{-1}(x)}{x^{3}}\textrm{d}x$ as a power series?

Question

anonymous · Answer

For starters, let's integrate:
$$\int\frac{x-\arctan x}{x^3}\,dx=\int\frac{dx}{x^2}-\int\frac{\arctan x}{x^3}\,dx$$
For the second integral, integrate by parts with
$$\begin{matrix}
u=\arctan x&&&dv=\dfrac{dx}{x^3}\
du=\dfrac{dx}{1+x^2}&&&v=-\dfrac{1}{2x^2}
\end{matrix}$$
So you have
$$\int\frac{dx}{x^2}-\left(-\frac{\arctan x}{2x^2}+\frac{1}{2}\int\frac{dx}{x^2(1+x^2)}\right)$$
$$-\frac{1}{x}+\frac{\arctan x}{2x^2}-\frac{1}{2}\int\frac{dx}{x^2(1+x^2)}$$
The remaining integral can be dealt with via partial fraction decomposition - I'll leave the work to you in verifying that the above is equivalent to
$$-\frac{1}{2x}+\frac{\arctan x}{2x^2}+\frac{1}{2}\arctan x$$
Now let's worry about the power series representation.

anonymous · Answer

Let's consider the derivative of $\arctan x$:
$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}$$
For $|-x^2|=|x|^2<1$, or simply $|x|<1$, the above has the geometric series representation,
$$\frac{d}{dx}\arctan x=\sum_{k=0}^\infty (-x^2)^k=\sum_{k=0}^\infty (-1)^kx^{2k}$$
Take the antiderivative will give us the power series for $\arctan x$:
$$\begin{align*}\int\frac{d}{dx}\arctan x\,dx&=\int\sum_{k=0}^\infty (-1)^kx^{2k}\,dx\\
\arctan x&=\int\sum_{k=0}^\infty (-1)^kx^{2k}\,dx\\
&=\sum_{k=0}^\infty (-1)^k\int x^{2k}\,dx\\
&=\sum_{k=0}^\infty (-1)^k\left(\frac{x^{2k+1}}{2k+1}+C\right)\end{align*}$$
We know that for $x=0$, we have $\arctan x=0$, so $C$ must also be $0$. (Alternatively, we can reason that if $C$ were non-zero, the series would not converge, so it must be zero.)

anonymous · Answer

So you can finally replace $\arctan x$ in original integral result and you'll have your very own power series for $\displaystyle\int\frac{x-\arctan x}{x^3}\,dx$.