Evaluate the integral: I = integral (5x+2)/(x²+2x+10) dx
Evaluate the integral: $$ I = \int \frac{5x+2}{x^2+2x+10}\,dx. $$
Explanation of Method
Completing the square in the denominator and splitting the numerator into a derivative-matching piece and a constant reduces the integral to a logarithm plus a standard arctangent integral.
- Complete the square for $x^2+2x+10$.
- Rewrite the numerator as $5(x+1)-3$.
- Split and integrate each piece.
Step 1 · Complete the square
$$ x^2+2x+10 = (x+1)^2+9. $$
Step 2 · Rewrite the numerator
$$ 5x+2 = 5(x+1) - 3. $$
Step 3 · Split and integrate
$$ I = \int \frac{5(x+1)}{(x+1)^2+9}\,dx - \int \frac{3}{(x+1)^2+9}\,dx. $$
For the first piece, let $v=(x+1)^2+9$, $dv=2(x+1)\,dx$:
$$ \int \frac{5(x+1)}{(x+1)^2+9}\,dx = \frac52\ln\left(x^2+2x+10\right). $$
For the second piece, using the standard formula $\int dv/(v^2+a^2)=\tfrac1a\arctan(v/a)$ with $a=3$:
$$ \int \frac{3}{(x+1)^2+9}\,dx = \arctan\left(\frac{x+1}{3}\right). $$
Step 4 · Combine
$$ I = \frac52\ln\left(x^2+2x+10\right) - \arctan\left(\frac{x+1}{3}\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $5(x+1)/[(x+1)^2+9]$ | $\tfrac52\ln(x^2+2x+10)$ |
| $3/[(x+1)^2+9]$ | $\arctan\!\left(\tfrac{x+1}3\right)$ |
Additional Notes
- The integrand is defined for all real $x$, since $x^2+2x+10>0$ everywhere (discriminant $4-40=-36$).
- This combines the two standard techniques from Problems 209 and 212's neighbors: derivative-matching for the log part and completing the square for the arctangent part, applied together in a single problem where the numerator isn't already proportional to the denominator's derivative.
- The result can be checked by differentiating $\tfrac52\ln(x^2+2x+10)-\arctan\!\left(\tfrac{x+1}3\right)$: this gives $\tfrac{5(x+1)}{x^2+2x+10}-\tfrac{3}{x^2+2x+10}=\tfrac{5x+5-3}{x^2+2x+10}=\tfrac{5x+2}{x^2+2x+10}$, matching the integrand.
Result
The solution to the integral is: