91Ó°ÊÓ

Consider the integral \( \int \sqrt{3 + 2x - x^2} \, dx \). We complete the square inside the square root: \[ 3 + 2x - x^2 = -(x^2 - 2x - 3) = -((x-1)^2 - 4) = -(x-1)^2 + 4 \].So the expression becomes \( \sqrt{4-(x-1)^2} \).This can be rewritten using the identity \( \sqrt{a^2 - u^2} \) as \( \sqrt{4-(x-1)^2} = \sqrt{2^2 - (x-1)^2} \), suggesting a trigonometric substitution.Use \( x-1 = 2\sin\theta \), so \( dx = 2\cos\theta \, d\theta \), and the integral becomes: \[ \int 2 \cos\theta \cdot 2 \cos \theta \, d\theta = 4 \int \cos^2 \theta \, d\theta \].Use the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \), the integral becomes: \[ 4 \left( \int \frac{1}{2} \, d\theta + \int \frac{\cos 2\theta}{2} \, d\theta \right) = 2\theta + 2 \int \cos 2\theta \, d\theta \].The second integral evaluates to\( \sin 2\theta \), thus:\[ 2\theta + \sin 2\theta + C \].Returning to \( x \) using \( \sin\theta = \frac{x-1}{2} \), we find \( \theta = \sin^{-1}\left(\frac{x-1}{2}\right) \). The final result is:\[ 2 \sin^{-1}\left(\frac{x-1}{2}\right) + \sin (2\sin^{-1}\left(\frac{x-1}{2}\right)) + C \].Simplify using the double angle identity for sine to complete the integral for problem (a).

For \( \int \frac{dx}{\sqrt{x^2 - 6x + 5}} \), start by completing the square: \[ x^2 - 6x + 5 = (x-3)^2 - 4 \].Therefore, the integral becomes \( \int \frac{dx}{\sqrt{(x-3)^2 - 2^2}} \), suggesting a hyperbolic substitution.Using \( x-3 = 2\cosh u \), thus \( dx = 2\sinh u \, du \), the integral simplifies to \[ \int \frac{2\sinh u}{2\sinh u} \, du = \int du = u + C \].Convert back to \( x \) with \( u = \cosh^{-1}\left(\frac{x-3}{2}\right) \), leading to:\[ \cosh^{-1}\left(\frac{x-3}{2}\right) + C \].

Consider \( \int \frac{dx}{\sqrt{x^2 - 4x + 8}} \). Complete the square in the denominator: \[ x^2 - 4x + 8 = (x-2)^2 + 4 \].This suggests the substitution \( x-2 = 2\tan \theta \), meaning \( dx = 2\sec^2 \theta \, d\theta \), the integral becomes: \[ \int \frac{2\sec^2 \theta}{\sqrt{4\tan^2 \theta + 4}} \, d\theta = \int \frac{2\sec^2 \theta}{2\sec \theta} \, d\theta = \int \sec \theta \, d\theta \].The integral \( \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C \).Return to \( x \) using \( \sec \theta = \frac{x}{2} \) and \( \tan \theta = \frac{x-2}{2} \). The result is:\[ \ln \left| \frac{x}{2} + \frac{x-2}{2} \right| + C \]. Note that this further simplifies within the specific conditions.

Address \( \int \frac{x+3}{\sqrt{x^2+4x+13}} \, dx \). First complete the square: \[ x^2 + 4x + 13 = (x+2)^2 + 9 \].Use the substitution \( x+2 = 3\sinh t \), \so \( dx = 3\cosh t \, dt \).The integral thus transforms to \[ \int \frac{3\sinh t + 1}{\sqrt{9\sinh^2 t + 9}} \cdot 3\cosh t \, dt \].Rewriting: \[ \int (9\sinh^2 t + 3\sinh t)\, dt \]. Expanding and solving each part gives hyperbolic sine and cosine primitives. Using identity methods, the decomposed result will lead to a logarithmic form based on simplification and integration processes.Once integrated, translate results back via hyperbolic transformations of \( x \) for the final form.

Given \( \int x \sqrt{3 + 2x - x^2} \, dx \), recognize that this extends problem (a).Reusing completed square \( 3 + 2x - x^2 = 4 - (x-1)^2 \) base from step 1, apply substitution \( x - 1 = 2\sin \theta \).Substitution becomes \( x = 2\sin \theta + 1 \), thus leading to the integral:\[ \int (2 \sin \theta + 1) \cdot 2 \cos \theta \cdot 2 \cos \theta \, d\theta \].Expanding yields the expression \( 4\int (2 \sin \theta \cos^2 \theta + \cos^2 \theta) \, d\theta \).Apply known integrals and identities (half-angle formulas) concluding with basis \( \theta \) applied back to \( x \), synthesizing final integration output.