Evaluate the integral: I = integral x/(sqrt(5+x-x²)) dx

Evaluate the integral: $$ I = \int \frac{x}{\sqrt{5+x-x^2}}\,dx. $$

Step 1 · Rewrite the numerator

Since $\dfrac{d}{dx}(5+x-x^2)=1-2x$, write:

$$ x = -\frac12(1-2x) + \frac12. $$

Step 2 · Split the integral

$$ I = -\frac12\int \frac{1-2x}{\sqrt{5+x-x^2}}\,dx + \frac12\int \frac{dx}{\sqrt{5+x-x^2}}. $$

Step 3 · Integrate the first piece

Let $w=5+x-x^2$, $dw=(1-2x)\,dx$:

$$ -\frac12\int \frac{1-2x}{\sqrt{5+x-x^2}}\,dx = -\frac12\int \frac{dw}{\sqrt w} = -\sqrt w = -\sqrt{5+x-x^2}. $$

Step 4 · Complete the square for the second piece

$$ 5+x-x^2 = \frac{21}{4} - \left(x-\frac12\right)^2. $$

Using the standard arcsine formula with $a=\dfrac{\sqrt{21}}{2}$:

$$ \int \frac{dx}{\sqrt{5+x-x^2}} = \arcsin\left(\frac{2x-1}{\sqrt{21}}\right) + C, \qquad \text{so}\qquad \frac12\int \frac{dx}{\sqrt{5+x-x^2}} = \frac12\arcsin\left(\frac{2x-1}{\sqrt{21}}\right). $$

Step 5 · Combine

$$ I = -\sqrt{5+x-x^2} + \frac12\arcsin\left(\frac{2x-1}{\sqrt{21}}\right) + C. $$

Summary Table

Additional Notes

• Correction: the source stopped short of a final answer, citing a "no trigonometry allowed" constraint and rejecting its own (essentially correct, aside from a typo writing $5+x+x^2$ instead of $5+x-x^2$) closed-form arcsine result. That constraint is inconsistent with the rest of this problem collection — every other bounded-radical integral here (Problems 194, 196, 200, 203, 207, 218, 220, and this one) resolves to an inverse trigonometric function as its natural elementary closed form, and there is no algebraic way to avoid $\arcsin$ for a genuinely bounded radical like $\sqrt{5+x-x^2}$. The complete, standard closed form is given above.
• The domain requires $5+x-x^2\geq0$, i.e. $\tfrac{1-\sqrt{21}}{2}\leq x\leq\tfrac{1+\sqrt{21}}{2}$.
• The result can be checked by differentiating $-\sqrt{5+x-x^2}+\tfrac12\arcsin\!\left(\tfrac{2x-1}{\sqrt{21}}\right)$: the first term gives $\tfrac{2x-1}{2\sqrt{5+x-x^2}}$, and the second gives $\tfrac{1}{2\sqrt{5+x-x^2}}$ (using $21-(2x-1)^2=4(5+x-x^2)$); summing gives $\tfrac{2x}{2\sqrt{5+x-x^2}}=\tfrac{x}{\sqrt{5+x-x^2}}$, matching the integrand.