Evaluate the integral: I = integral (xe^(arctan

Evaluate the integral: $$ I = \int \frac{xe^{\arctan x}}{(1+x^2)^{3/2}}\,dx. $$

Step 1 · Substitution

Let $t=\arctan x$, so $x=\tan t$ and $dx=\sec^2t\,dt$, and $1+x^2=\sec^2t$. Substituting:

$$ I = \int \frac{\tan t\,e^t}{(\sec^2t)^{3/2}}\cdot\sec^2t\,dt = \int \frac{\tan t\,e^t}{\sec^3t}\cdot\sec^2t\,dt = \int \tan t\,e^t\cdot\frac{1}{\sec t}\,dt. $$

Step 2 · Simplify

Since $\tan t/\sec t=\sin t$:

$$ I = \int \sin t\,e^t\,dt. $$

Step 3 · Apply the self-referential technique

This is the same integral solved in Problem 162: applying integration by parts twice and solving algebraically gives:

$$ I = \frac{e^t}{2}\left(\sin t-\cos t\right) + C. $$

Step 4 · Substitute back

Since $t=\arctan x$, the right-triangle relationships give $\sin t=\dfrac{x}{\sqrt{1+x^2}}$ and $\cos t=\dfrac{1}{\sqrt{1+x^2}}$. Substituting:

$$ I = \frac{e^{\arctan x}}{2}\left(\frac{x}{\sqrt{1+x^2}} - \frac{1}{\sqrt{1+x^2}}\right) + C = \frac{e^{\arctan x}}{2}\cdot\frac{x-1}{\sqrt{1+x^2}} + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$.
• Simplifying $\tan t\,e^t/\sec t$ to $\sin t\,e^t$ before applying integration by parts avoids ever having to differentiate the messier original composite expression directly — recognizing the substitution collapses to the already-solved pattern from Problem 162 saves significant work.
• The result can be checked by differentiating $\tfrac{e^{\arctan x}}{2}\cdot\tfrac{x-1}{\sqrt{1+x^2}}$ and confirming it returns $xe^{\arctan x}/(1+x^2)^{3/2}$.