How do you integrate #(x*arctan(x))/(1+x^2)^2#?

Question

Ezra Adams · Accepted Answer

#= ( x + (x^2 -1) arctan x )/(4(x^2+1)) + C#

#int (x arctan(x))/(1+x^2)^2 dx#

#= int arctan(x) (x )/(1+x^2)^2 dx#

#= int arctan(x) ( (- 1/2 )/(1+x^2) )^' dx#

#= -1/2 int arctan(x) ( (1)/(1+x^2) )^' dx#

which by IBP

#= -1/2 ( arctan(x) (1)/(1+x^2) - int ( arctan(x) )^' (1)/(1+x^2) dx )#

#= -1/2 ( arctan(x) (1)/(1+x^2) - color(blue)( int 1/(1+x^2)^2 dx) ) qquad square#

for the blue bit we say that #x = tan xi, dx = sec^2 xi \ d xi#

So we have

#int 1/(1+tan^2 xi)^2 sec^2 xi \ d xi = int cos^2 xi \ d xi#

for this we use the double angle formula: #cos 2 gamma = 2 cos^2 gamma - 1# giving us:

#1/2 int cos 2 xi + 1 \ d xi#

#= 1/2 ( 1/2 sin 2 xi + xi ) + C#

#= 1/2 ( sin xi cos xi + xi ) + C qquad triangle#

and if #tan xi = x#, then #sin xi = x/sqrt(x^2 +1)# and #cos xi = 1/sqrt(x^2 +1)#, just draw the right-angled triangle to see.

so #triangle # becomes

#= 1/2 ( x/(x^2 +1) + arctan x ) + C qquad triangle#

So when we pop it back into #square# we get

#= -1/2 ( arctan(x) (1)/(1+x^2) - (1/2 ( x/(x^2 +1) + arctan x ) + C ) #

#= -1/2 arctan(x) (1)/(x^2+1) + 1/4 x/(x^2 +1) + 1/4 arctan x + C#

Tidying up

#=1/4 1/(x^2+1) ( -2 arctan(x) + x + (x^2 +1) arctan x ) + C#

#=1/4 1/(x^2+1) ( x + (x^2 -1) arctan x ) + C#

#= ( x + (x^2 -1) arctan x )/(4(x^2+1)) + C#

Christopher Allers · Answer

undefined User is interested in receiving answers to questions without any irrelevant information or introduction words.To integrate $\frac{x\cdot \text{arctan}(x)}{(1+x^2)^2}$, you can use integration by parts. Let $u = \text{arctan}(x)$ and $dv = \frac{x}{(1+x^2)^2}dx$. Then, $du = \frac{1}{1+x^2}dx$ and $v = -\frac{1}{2(1+x^2)}$. Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, you get:

$$
\begin{aligned}
\int \frac{x\cdot \text{arctan}(x)}{(1+x^2)^2} \, dx &= -\frac{x\cdot \text{arctan}(x)}{2(1+x^2)} - \int -\frac{1}{2(1+x^2)} \cdot \frac{1}{1+x^2} \, dx \
&= -\frac{x\cdot \text{arctan}(x)}{2(1+x^2)} + \frac{1}{2} \int \frac{dx}{(1+x^2)^2}.
\end{aligned}
$$

To integrate $\frac{1}{(1+x^2)^2}$, you can use the substitution method. Let $u = 1+x^2$, then $du = 2x \, dx$, and the integral becomes:

$$
\begin{aligned}
\frac{1}{2} \int \frac{dx}{(1+x^2)^2} &= \frac{1}{2} \int \frac{du}{2u^2} \
&= \frac{1}{4} \int u^{-2} \, du \
&= \frac{1}{4} \cdot \frac{u^{-1}}{-1} + C \
&= -\frac{1}{4u} + C \
&= -\frac{1}{4(1+x^2)} + C,
\end{aligned}
$$

where $C$ is the constant of integration. Substituting this back into the previous expression, you get the final result:

$$
\int \frac{x\cdot \text{arctan}(x)}{(1+x^2)^2} \, dx = -\frac{x\cdot \text{arctan}(x)}{2(1+x^2)} - \frac{1}{4(1+x^2)} + C,
$$

where $C$ is the constant of integration.