Evaluate the integral: I = integral (x^6)/(sqrt(x²+1)) dx
Evaluate the integral: $$ I = \int \frac{x^6}{\sqrt{x^2+1}}\,dx. $$
Explanation of Method
Writing $x^6=(x^2+1-1)^3$ and expanding via the binomial theorem turns the integrand into a sum of $(x^2+1)^{n/2}$ terms, each handled by the standard reduction formula $\int(x^2+1)^{n/2}dx=\tfrac{x(x^2+1)^{n/2}}{n+1}+\tfrac n{n+1}\int(x^2+1)^{(n-2)/2}dx$.
- Expand $x^6$ in terms of $(x^2+1)$.
- Apply the reduction formula to each power.
- Combine all terms.
Step 1 · Expand
$$ x^6 = \left[(x^2+1)-1\right]^3 = (x^2+1)^3 - 3(x^2+1)^2 + 3(x^2+1) - 1. $$
$$ I = \int(x^2+1)^{5/2}dx - 3\int(x^2+1)^{3/2}dx + 3\int(x^2+1)^{1/2}dx - \int\frac{dx}{\sqrt{x^2+1}}. $$
Step 2 · Standard base cases
$$ \int\frac{dx}{\sqrt{x^2+1}}=\ln\left(x+\sqrt{x^2+1}\right), \qquad \int\sqrt{x^2+1}\,dx=\frac{x\sqrt{x^2+1}}2+\frac12\ln\left(x+\sqrt{x^2+1}\right). $$
Step 3 · Apply the reduction formula twice more
$$ \int(x^2+1)^{3/2}dx = \frac{x(x^2+1)^{3/2}}4+\frac{3x\sqrt{x^2+1}}8+\frac38\ln\left(x+\sqrt{x^2+1}\right), $$
$$ \int(x^2+1)^{5/2}dx = \frac{x(x^2+1)^{5/2}}6+\frac{5x(x^2+1)^{3/2}}{24}+\frac{5x\sqrt{x^2+1}}{16}+\frac5{16}\ln\left(x+\sqrt{x^2+1}\right). $$
Step 4 · Combine
Substituting into Step 1 and collecting coefficients of each term:
Summary Table
| Term | Coefficient |
|---|---|
| $x(x^2+1)^{5/2}$ | $1/6$ |
| $x(x^2+1)^{3/2}$ | $-13/24$ |
| $x\sqrt{x^2+1}$ | $11/16$ |
| $\ln(x+\sqrt{x^2+1})$ | $-5/16$ |
Additional Notes
- The domain is all real $x$.
- The result was verified by differentiation at $x=1$ and $x=2$: the derivative matches the integrand $x^6/\sqrt{x^2+1}$ exactly at both points.
Result
The solution to the integral is: