Evaluate the integral: I = integral x^7sqrt(1+x²) dx

Evaluate the integral: $$ I = \int x^7\sqrt{1+x^2}\,dx. $$

Step 1 · Substitute

Let $u=1+x^2$, so $du=2x\,dx$ and $x^6=(u-1)^3$:

$$ I = \int x^6\sqrt{1+x^2}\cdot x\,dx = \frac12\int(u-1)^3\sqrt u\,du. $$

Step 2 · Expand and integrate

Expanding $(u-1)^3=u^3-3u^2+3u-1$ and multiplying by $u^{1/2}$:

$$ \frac12\int\left(u^{7/2}-3u^{5/2}+3u^{3/2}-u^{1/2}\right)du = \frac{u^{9/2}}9 - \frac{3u^{7/2}}7 + \frac{3u^{5/2}}5 - \frac{u^{3/2}}3 + C. $$

Step 3 · Back-substitute

With $u=1+x^2$:

$$ I = \frac{(1+x^2)^{9/2}}9 - \frac{3(1+x^2)^{7/2}}7 + \frac{3(1+x^2)^{5/2}}5 - \frac{(1+x^2)^{3/2}}3 + C. $$

Summary Table

Additional Notes

• The domain is all real $x$, since $1+x^2>0$ everywhere.
• The result was verified symbolically: differentiating the boxed expression simplifies exactly to $x^7\sqrt{1+x^2}$, confirmed numerically at $x=1$.