Evaluate the integral: I = integral dx/(sqrt(x+x²))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{x+x^2}}. $$
Explanation of Method
Shifting to $u=x+1$ and then substituting $u=t^2$ reduces the radical to a standard $\int dt/\sqrt{t^2-1}$ form, whose logarithmic antiderivative simplifies neatly back in terms of $x$.
- Shift using $u=x+1$.
- Substitute $u=t^2$.
- Apply the standard result $\int dt/\sqrt{t^2-1}=\ln|t+\sqrt{t^2-1}|$.
- Substitute back to $x$.
Step 1 · Shift the variable
Let $x=u-1$, $dx=du$. Then:
$$ x+x^2 = (u-1)+(u-1)^2 = u-1+u^2-2u+1 = u^2-u = u(u-1). $$
$$ I = \int \frac{du}{\sqrt{u(u-1)}}. $$
Step 2 · Substitute $u=t^2$
With $du=2t\,dt$ and $u-1=t^2-1$:
$$ I = \int \frac{2t\,dt}{\sqrt{t^2(t^2-1)}} = \int \frac{2t\,dt}{t\sqrt{t^2-1}} = 2\int \frac{dt}{\sqrt{t^2-1}}. $$
Step 3 · Apply the standard result
$$ 2\int \frac{dt}{\sqrt{t^2-1}} = 2\ln\left|t+\sqrt{t^2-1}\right| + C. $$
Step 4 · Substitute back
With $t=\sqrt u$ and then $u=x+1$:
$$ I = 2\ln\left|\sqrt{u}+\sqrt{u-1}\right| + C = 2\ln\left|\sqrt{x+1}+\sqrt{x}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Shift | $u=x+1$ |
| Substitution | $u=t^2$ |
| Reduced integral | $2\int dt/\sqrt{t^2-1}$ |
Additional Notes
- The domain requires $x+x^2\geq0$, i.e. $x\leq-1$ or $x\geq0$.
- This double-substitution route (shift, then $u=t^2$) is equivalent to directly completing the square on $x^2+x$ and applying $\int du/\sqrt{u^2-a^2}=\ln|u+\sqrt{u^2-a^2}|$; both give antiderivatives that differ only by an additive constant, since $2\ln(\sqrt{x+1}+\sqrt x)=\ln\!\left[(\sqrt{x+1}+\sqrt x)^2\right]=\ln\!\left[2x+1+2\sqrt{x^2+x}\right]$.
- The result can be checked by differentiating $2\ln(\sqrt{x+1}+\sqrt x)$: this gives $2\cdot\dfrac{\tfrac{1}{2\sqrt{x+1}}+\tfrac{1}{2\sqrt x}}{\sqrt{x+1}+\sqrt x}=\dfrac{1/\sqrt{x+1}+1/\sqrt x}{\sqrt{x+1}+\sqrt x}=\dfrac{(\sqrt x+\sqrt{x+1})/\sqrt{x(x+1)}}{\sqrt{x+1}+\sqrt x}=\dfrac{1}{\sqrt{x(x+1)}}=\dfrac{1}{\sqrt{x^2+x}}$, matching the integrand.
Result
The solution to the integral is: