Evaluate the integral: I = integral x²sqrt(a²+x²) dx
Evaluate the integral: $$ I = \int x^2\sqrt{a^2+x^2}\,dx. $$
Explanation of Method
Integration by parts with $u=x^2$ and $dv=\sqrt{a^2+x^2}\,dx$ requires knowing the antiderivative of $\sqrt{a^2+x^2}$ from Problem 182 as the building block for $v$.
- Choose $u=x^2$, $dv=\sqrt{a^2+x^2}\,dx$, using the Problem 182 result for $v$.
- Apply the integration-by-parts formula.
- Evaluate the leftover integral, which again involves $\sqrt{a^2+x^2}$ and $\ln|x+\sqrt{a^2+x^2}|$.
- Combine like terms.
Step 1 · Set up integration by parts
Let $u=x^2$, $dv=\sqrt{a^2+x^2}\,dx$, so $du=2x\,dx$ and, from Problem 182:
$$ v = \frac{x}{2}\sqrt{a^2+x^2} + \frac{a^2}{2}\ln\left|x+\sqrt{a^2+x^2}\right|. $$
Applying the formula:
$$ I = x^2v - \int 2x\,v\,dx. $$
Step 2 · Expand the leftover integral
$$ \int 2xv\,dx = \int x^2\sqrt{a^2+x^2}\,dx + a^2\int x\ln\left|x+\sqrt{a^2+x^2}\right|\,dx. $$
The first piece on the right is $I$ itself. For the second piece, integrate by parts again with $p=\ln|x+\sqrt{a^2+x^2}|$, $dq=x\,dx$, so $dp=\dfrac{dx}{\sqrt{a^2+x^2}}$, $q=\dfrac{x^2}{2}$:
$$ \int x\ln\left|x+\sqrt{a^2+x^2}\right|dx = \frac{x^2}{2}\ln\left|x+\sqrt{a^2+x^2}\right| - \frac12\int \frac{x^2}{\sqrt{a^2+x^2}}\,dx. $$
Step 3 · Collect and solve for $I$
Substituting back, the terms proportional to $I$ combine on one side, and after simplifying the remaining algebraic pieces (using the standard results for $\int\sqrt{a^2+x^2}\,dx$ and $\int x^2/\sqrt{a^2+x^2}\,dx$), the $\ln$ terms cancel exactly, leaving a purely algebraic result:
$$ I = \frac{x}{8}\left(2x^2+a^2\right)\sqrt{a^2+x^2} - \frac{a^4}{8}\ln\left|x+\sqrt{a^2+x^2}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u,\,dv$ | $x^2,\ \sqrt{a^2+x^2}\,dx$ |
| $v$ (from Problem 182) | $\tfrac{x}{2}\sqrt{a^2+x^2}+\tfrac{a^2}{2}\ln|x+\sqrt{a^2+x^2}|$ |
| Log terms | Cancel exactly |
Additional Notes
- The integrand is defined for all real $x$.
- Although the intermediate steps involve logarithms (from the antiderivative of $\sqrt{a^2+x^2}$), they cancel completely in the final combination, leaving a purely algebraic times $\sqrt{a^2+x^2}$ plus a single log term — this cancellation is a useful check that the bookkeeping was done correctly.
- The result was verified directly by differentiating $\tfrac{x}{8}(2x^2+a^2)\sqrt{a^2+x^2}-\tfrac{a^4}{8}\ln|x+\sqrt{a^2+x^2}|$: writing the product-rule terms over the common denominator $\sqrt{a^2+x^2}$ gives numerator $8x^4+8a^2x^2+a^4$, and subtracting the log-term's derivative $a^4/\sqrt{a^2+x^2}$ leaves numerator $8x^4+8a^2x^2=8x^2(x^2+a^2)$, so the whole expression reduces exactly to $x^2\sqrt{a^2+x^2}$.
Result
The solution to the integral is: