Evaluate the integral: I = integral (x²+1)/(x

Evaluate the integral: $$ I = \int \frac{x^2+1}{x\sqrt{x^2+x^4+1}}\,dx. $$

Explanation of Method

Dividing numerator and denominator by $x^2$ turns the integrand into a function of $u=x-1/x$, since $du=(1+1/x^2)\,dx$ matches the numerator exactly and $x^2+1/x^2+1=u^2+3$.

Divide by $x^2$.
Substitute $u=x-1/x$.
Integrate the standard radical form.
Back-substitute.

Step 1 · Divide by $x^2$

For $x>0$:

$$ I = \int\frac{1+1/x^2}{\sqrt{1+x^2+1/x^2}}\,dx. $$

Step 2 · Substitute

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

$$ I = \int\frac{du}{\sqrt{u^2+3}}. $$

Step 3 · Integrate

$$ I = \ln\left|u+\sqrt{u^2+3}\right| + C. $$

Step 4 · Back-substitute

With $u=x-1/x=(x^2-1)/x$ and $\sqrt{u^2+3}=\sqrt{x^2+x^4+1}/x$ (for $x>0$):

Summary Table

Substitution	Effect
$u=x-1/x$	reduces to $\int du/\sqrt{u^2+3}$

Additional Notes

The domain excludes $x=0$.
The result was verified by differentiation at $x=1$ and $x=2$: the derivative matches the integrand $(x^2+1)/[x\sqrt{x^2+x^4+1}]$ exactly at both points.

Result

The solution to the integral is:

$$ I = \ln\left|\frac{x^2-1+\sqrt{x^2+x^4+1}}x\right| + C. $$