Integral with partial fractions

Question

I'm trying to calculate the following integral:

$$I=\int \frac{\arctan(x)}{x^4}dx$$

My steps so far are:

Per partes:

$$\frac{-\arctan(x)}{3x^3}+\int{\frac{1}{1+x^2} \frac{1}{3x^3}}dx =\frac{-\arctan(x)}{3x^3}+\frac{1}{3}\int{\frac{1}{1+x^2} \frac{1}{x^3}}dx$$

and now I want to do partial fractions. However, with this integral, I fail to do partial fractions. Could you help me?

Thanks

Answer 1

A rational function $P(x)/Q(x)$ can be rewritten using Partial Fraction Decomposition:

$$ \frac{P(x)}{Q(x)} = \frac{A_1}{a\,x + b} + \dots + \frac{A_2\,x + B_2}{a\,x^2 + b\,x + c} + \dots $$

where for each factor of $Q(x)$ of the form $(a\,x + b)^m$ introduce terms:

$$ \frac{A_1}{a\,x + b} + \frac{A_2}{(a\,x + b)^2} + \dots + \frac{A_m}{(a\,x + b)^m} $$

and for each factor of $Q(x)$ of the form $\left(a\,x^2 + b\,x + c\right)^m$ introduce terms:

$$ \frac{A_1\,x + B_1}{a\,x^2 + b\,x + c} + \frac{A_2\,x + B_2}{\left(a\,x^2 + b\,x + c\right)^2} + \dots + \frac{A_m\,x + B_m}{\left(a\,x^2 + b\,x + c\right)^m}\,. $$

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In light of all this, you have:

$$ \frac{1}{x^3\left(x^2+1\right)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x^3} + \frac{A_4\,x + B_4}{x^2 + 1} $$

i.e.

$$ \frac{1}{x^3\left(x^2+1\right)} = \frac{\left(A_1 + A_4\right)x^4 + \left(A_2 + B_4\right)x^3 + \left(A_1 + A_3\right)x^2 + A_2\,x + A_3}{x^3\left(x^2+1\right)} $$

which turns out to be an identity if and only if:

$$ \begin{cases} A_1 + A_4 = 0 \\ A_2 + B_4 = 0 \\ A_1 + A_3 = 0 \\ A_2 = 0 \\ A_3 = 1 \end{cases} \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} A_1 = -1 \\ A_2 = 0 \\ A_3 = 1 \\ A_4 = 1 \\ B_4 = 0 \end{cases} $$

from which what you want:

$$ \frac{1}{x^3\left(x^2+1\right)} = -\frac{1}{x} + \frac{1}{x^3} + \frac{x}{x^2+1}\,. $$

Answer 2

Or more simply:

$$A=\frac{1}{1+x^2}*\frac{1}{x^3}$$ $$A=-\frac 1 x \left (\frac{1}{1+x^2}-\frac{x}{x^3} \right )$$ $$A=-\frac{1}{x(1+x^2)}+\frac{1}{x^3}$$

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$$B=\frac{1}{x(1+x^2)}=\frac{1}{x}-\frac x {(1+x^2)}$$ We have that: $$I=\int \left \{\frac x {(1+x^2)}-\frac{1}{x}+\frac{1}{x^3} \right \}dx$$

Answer 3

Without partial fractions $$ \begin{aligned} I=\int \frac{\arctan x}{x^4} d x & =-\frac{1}{3} \int \arctan x d\left(\frac{1}{x^3}\right) \\ & \stackrel{IBP}{=} -\frac{\arctan x}{3 x^3}+\frac{1}{3} \underbrace{\int \frac{1}{x^3\left(1+x^2\right)}}_J dx \end{aligned} $$ Letting $y=\frac{1}{x} $ yields $$ \begin{aligned} J & =\int \frac{1}{\frac{1}{y^3}\left(1+\frac{1}{y^2}\right)}\left(-\frac{d y}{y^2}\right) \\ & =-\int \frac{y^3}{y^2+1} d y \\ & =-\int \frac{y\left(y^2+1\right)-y}{y^2+1} d y \\ & =-\frac{y^2}{2}+\frac{1}{2} \ln \left(y^2+1\right)+C \end{aligned} $$ Hence $$ I=\frac{1}{6}\left[-\frac{2 \arctan x}{x^3}-\frac{1}{x^2}+\ln \left(1+x^2\right)-2 \ln |x|\right]+C $$

Answer 4

Alternatively, substitute $u=\arctan(x)$, then integrate by parts.

$$\int \frac{\arctan(x)}{x^4} \, dx = \int u \cot^2(u)\csc^2(u) \, du = UV-\int V\,dU$$

where $U=u$ and $dV=\cot^2(u)\csc^2(u)\,du$.

$$\begin{align*} \int \frac{\arctan(x)}{x^4} \, dx &= -\frac13 u\cot^3(u) + \frac13 \int \cot^3(u) \, du \\ &= -\frac13 u\cot^3(u) - \frac16 \csc^2(u) -\frac13 \ln\lvert\sin(u)\rvert + C \\ &= -\frac{\arctan(x)}{3x^3} - \frac{1+\color{red}{x^2}}{6x^2} - \frac13 \ln\frac{\lvert x\rvert}{\sqrt{1+x^2}} + C \end{align*}$$

and the red term can be absorbed into $C$.