Evaluate the integral: I = integral dx/((1+x²)^(3/2))

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

Explanation of Method

The trig substitution $x=\tan\theta$ collapses $(1+x^2)^{3/2}=\sec^3\theta$ against the $\sec^2\theta\,d\theta$ from $dx$, reducing the integral to $\int\cos\theta\,d\theta$.

Substitute $x=\tan\theta$.
Integrate $\cos\theta$.
Convert back to $x$.

Step 1 · Substitute

Let $x=\tan\theta$, so $dx=\sec^2\theta\,d\theta$ and $1+x^2=\sec^2\theta$:

$$ I = \int\frac{\sec^2\theta\,d\theta}{\sec^3\theta} = \int\cos\theta\,d\theta. $$

Step 2 · Integrate

$$ I = \sin\theta + C. $$

Step 3 · Convert back to $x$

From a right triangle with opposite $x$, adjacent $1$, hypotenuse $\sqrt{1+x^2}$: $\sin\theta=x/\sqrt{1+x^2}$.

Summary Table

Quantity	Value
$\sin\theta$	$x/\sqrt{1+x^2}$

Additional Notes

The domain is all real $x$.
The result was verified by differentiation at $x=1$: using the quotient rule, $\tfrac{d}{dx}\left[x/\sqrt{1+x^2}\right]=1/(1+x^2)^{3/2}$, matching the integrand exactly.

Result

The solution to the integral is:

$$ I = \frac x{\sqrt{1+x^2}} + C. $$