Evaluate the integral: I = integral (x dx)/((1 + x²)²)
Evaluate the integral: $$ I = \int \frac{x\,dx}{(1 + x^2)^2}. $$
Explanation of Method
Substituting $u=1+x^2$ turns the squared denominator into $u^2$, and since $du=2x\,dx$ matches the numerator up to a constant, the integral reduces to a simple negative-power-rule antiderivative.
- Substitute $u$ for the base of the denominator.
- Match $x\,dx$ to a constant multiple of $du$.
- Integrate the resulting negative power of $u$.
- Substitute back and add the constant of integration.
Step 1 · Set up the substitution
To solve this integral, we use the substitution method. Observing the denominator, we choose:
$$ u = 1 + x^2. $$
Now, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = 2x. $$
Rearranging, we get:
$$ du = 2x\,dx. $$
This allows us to express $x\,dx$ in terms of $du$:
$$ x\,dx = \frac{du}{2}. $$
Step 2 · Rewrite and integrate
Substituting into the integral:
$$ I = \int \frac{\frac{du}{2}}{u^2}. $$
Factor out the constant:
$$ I = \frac{1}{2}\int u^{-2}\,du. $$
The integral of $u^{-2}$ is:
$$ \int u^{-2}\,du = -\frac{1}{u}. $$
Thus, we obtain:
$$ I = \frac{1}{2}\left(-\frac{1}{u}\right) + C = -\frac{1}{2u} + C. $$
Step 3 · Substitute back
Finally, substituting back $u = 1+x^2$, we get:
$$ I = -\frac{1}{2(1+x^2)} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 1+x^2$ |
| $x\,dx$ in terms of $du$ | $\tfrac12 du$ |
| Antiderivative | $-\tfrac{1}{2u}$ |
Additional Notes
- The integrand is defined for all real $x$, since $1+x^2 \geq 1 > 0$ always.
- This is a power-rule integral in disguise, not a logarithmic one — since the exponent on $u$ is $-2$ rather than $-1$, the antiderivative is $-\tfrac{1}{2u}$, not a logarithm.
- The result can be checked by differentiating $-\dfrac{1}{2(1+x^2)}$ with the chain rule and confirming it returns $\dfrac{x}{(1+x^2)^2}$.
Result
The solution to the integral is: