Evaluate the integral: I = integral (sqrt(x²+2))/(x²+1) dx
Evaluate the integral: $$ I = \int \frac{\sqrt{x^2+2}}{x^2+1}\,dx. $$
Explanation of Method
Writing $x^2+2=(x^2+1)+1$ splits the fraction into $\sqrt{x^2+2}/(x^2+2)$-style pieces after multiplying by $\sqrt{x^2+2}/\sqrt{x^2+2}$, giving a standard radical integral plus an integral of $1/[(x^2+1)\sqrt{x^2+2}]$, which a $u=\sqrt{x^2+2}$ substitution followed by a trig substitution resolves into an arctangent.
- Split the integrand.
- Integrate the first piece.
- Integrate the second piece via substitution.
Step 1 · Split
Since $x^2+2=(x^2+1)+1$:
$$ \frac{\sqrt{x^2+2}}{x^2+1} = \frac{x^2+2}{\sqrt{x^2+2}(x^2+1)} = \frac1{\sqrt{x^2+2}} + \frac1{(x^2+1)\sqrt{x^2+2}}. $$
Step 2 · First piece
$$ \int\frac{dx}{\sqrt{x^2+2}} = \ln\left|x+\sqrt{x^2+2}\right| + C_1. $$
Step 3 · Second piece
Let $x=\tan\varphi$ was tried first, but a cleaner route uses $u=\sqrt{x^2+2}$ then $u=\sqrt2\sec\phi$, reducing $\int dx/[(x^2+1)\sqrt{x^2+2}]$ to $\int\cos\phi\,d\phi/(2-\cos^2\phi)$. With $w=\sin\phi$:
$$ \int\frac{dw}{1+w^2} = \arctan(w) + C_2 = \arctan\left(\frac x{\sqrt{x^2+2}}\right) + C_2. $$
Step 4 · Combine
$$ I = \ln\left|x+\sqrt{x^2+2}\right| + \arctan\left(\frac x{\sqrt{x^2+2}}\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $1/\sqrt{x^2+2}$ | $\ln|x+\sqrt{x^2+2}|$ |
| $1/[(x^2+1)\sqrt{x^2+2}]$ | $\arctan\left(x/\sqrt{x^2+2}\right)$ |
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 $\sqrt{x^2+2}/(x^2+1)$ exactly at both points.
Result
The solution to the integral is: