Evaluate the integral: I = integral dx/((x²+1)³)
Evaluate the integral: $$ I = \int \frac{dx}{(x^2+1)^3}. $$
Explanation of Method
Substituting $x=\tan\theta$ reduces the integral to $\int\cos^4\theta\,d\theta$, resolved by two applications of the power-reduction identity, then converted back to $x$ using $\sin2\theta$ and $\sin4\theta$ in terms of $\tan\theta$.
- Substitute $x=\tan\theta$.
- Reduce $\cos^4\theta$ using power-reduction identities twice.
- Integrate term by term.
- Substitute back, keeping both the $\sin2\theta$ and $\sin4\theta$ contributions.
Step 1 · Trigonometric substitution
Let $x=\tan\theta$, $dx=\sec^2\theta\,d\theta$, $x^2+1=\sec^2\theta$:
$$ I = \int \frac{\sec^2\theta\,d\theta}{\sec^6\theta} = \int \cos^4\theta\,d\theta. $$
Step 2 · Power-reduce $\cos^4\theta$
Using $\cos^2\theta=\tfrac{1+\cos2\theta}{2}$ twice:
$$ \cos^4\theta = \frac{3}{8} + \frac12\cos2\theta + \frac18\cos4\theta. $$
Step 3 · Integrate
$$ I = \frac{3\theta}{8} + \frac14\sin2\theta + \frac{1}{32}\sin4\theta + C. $$
Step 4 · Substitute back
With $\theta=\arctan x$, $\sin2\theta=\dfrac{2x}{1+x^2}$, and $\sin4\theta=2\sin2\theta\cos2\theta=\dfrac{4x(1-x^2)}{(1+x^2)^2}$:
$$ I = \frac38\arctan x + \frac14\cdot\frac{2x}{1+x^2} + \frac{1}{32}\cdot\frac{4x(1-x^2)}{(1+x^2)^2} + C. $$
Combining the two algebraic terms over a common denominator $8(1+x^2)^2$:
$$ \frac{x}{2(1+x^2)} + \frac{x(1-x^2)}{8(1+x^2)^2} = \frac{4x(1+x^2)+x(1-x^2)}{8(1+x^2)^2} = \frac{x(3x^2+5)}{8(1+x^2)^2}. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $x=\tan\theta$ |
| Reduced integral | $\int\cos^4\theta\,d\theta$ |
| Combined algebraic term | $x(3x^2+5)/[8(1+x^2)^2]$ |
Additional Notes
- Correction: the source's Step 6 correctly listed both the $\sin2\theta$ and $\sin4\theta$ back-substitution terms, but the "Step 7 Final Answer" silently dropped the entire $\sin4\theta$ contribution and also collapsed $\theta/4+\theta/8=3\theta/8$ down to just $\theta/4$, with no algebra shown to justify either change. The complete, correct combination keeps all three pieces, matching the well-known reduction formula $\int dx/(x^2+1)^3=\tfrac{x(3x^2+5)}{8(1+x^2)^2}+\tfrac38\arctan x+C$.
- The integrand is defined for all real $x$.
- The result was verified directly by differentiation: writing the quotient-rule derivative of $\tfrac{x(3x^2+5)}{8(1+x^2)^2}$ over the common denominator $8(1+x^2)^3$ gives numerator $-3x^4-6x^2+5$, and adding the arctangent term's derivative $\tfrac{3(1+x^2)^2}{8(1+x^2)^3}$ (numerator $3+6x^2+3x^4$) sums to a constant numerator of $8$, reducing the whole expression to exactly $1/(1+x^2)^3$.
Result
The solution to the integral is: