1. **Problem I:** Evaluate $$\int \frac{dx}{x \sqrt{x^2 - \pi}}$$ - Use substitution for integrals involving $$\sqrt{x^2 - a^2}$$. - Let $$x = \sqrt{\pi} \sec \theta$$, then $$dx = \sqrt{\pi} \sec \theta \tan \theta d\theta$$. - Substitute into the integral: $$\int \frac{\sqrt{\pi} \sec \theta \tan \theta d\theta}{\sqrt{\pi} \sec \theta \sqrt{\pi \sec^2 \theta - \pi}} = \int \frac{\sqrt{\pi} \sec \theta \tan \theta d\theta}{\sqrt{\pi} \sec \theta \sqrt{\pi (\sec^2 \theta - 1)}}$$ - Simplify inside the square root: $$\sqrt{\pi (\sec^2 \theta - 1)} = \sqrt{\pi \tan^2 \theta} = \sqrt{\pi} \tan \theta$$ - The integral becomes: $$\int \frac{\sqrt{\pi} \sec \theta \tan \theta d\theta}{\sqrt{\pi} \sec \theta \sqrt{\pi} \tan \theta} = \int \frac{\sqrt{\pi} \sec \theta \tan \theta d\theta}{\sqrt{\pi} \sec \theta \sqrt{\pi} \tan \theta}$$ - Cancel common factors: $$\int \frac{\cancel{\sqrt{\pi}} \cancel{\sec \theta} \cancel{\tan \theta} d\theta}{\cancel{\sqrt{\pi}} \cancel{\sec \theta} \cancel{\sqrt{\pi}} \cancel{\tan \theta}} = \int \frac{d\theta}{\sqrt{\pi}}$$ - So the integral is: $$\frac{1}{\sqrt{\pi}} \int d\theta = \frac{\theta}{\sqrt{\pi}} + C$$ - Back-substitute $$\theta = \sec^{-1} \left( \frac{x}{\sqrt{\pi}} \right)$$: $$\boxed{\frac{\sec^{-1} \left( \frac{x}{\sqrt{\pi}} \right)}{\sqrt{\pi}} + C}$$ 2. **Problem II:** Evaluate $$\int \frac{e^t}{1 + e^{2x}} dx$$ - The integral is with respect to $$x$$ but the numerator is $$e^t$$, which is independent of $$x$$. - Treat $$e^t$$ as a constant: $$e^t \int \frac{dx}{1 + e^{2x}}$$ - Substitute $$u = e^x$$, so $$du = e^x dx = u dx \Rightarrow dx = \frac{du}{u}$$. - The integral becomes: $$e^t \int \frac{1}{1 + u^2} \cdot \frac{du}{u}$$ - This is: $$e^t \int \frac{du}{u(1 + u^2)}$$ - Use partial fractions: $$\frac{1}{u(1+u^2)} = \frac{A}{u} + \frac{Bu + C}{1 + u^2}$$ - Solving gives $$A=1$$, $$B=-1$$, $$C=0$$. - So: $$\int \frac{du}{u(1+u^2)} = \int \left( \frac{1}{u} - \frac{u}{1+u^2} \right) du = \int \frac{1}{u} du - \int \frac{u}{1+u^2} du$$ - Integrate: $$\ln|u| - \frac{1}{2} \ln(1+u^2) + C$$ - Back-substitute $$u = e^x$$: $$e^t \left( x - \frac{1}{2} \ln(1 + e^{2x}) \right) + C$$ 3. **Problem III:** Evaluate $$\int \sqrt{x} \ln x dx$$ - Use integration by parts: - Let $$u = \ln x$$, $$dv = \sqrt{x} dx = x^{1/2} dx$$ - Then $$du = \frac{1}{x} dx$$, $$v = \frac{2}{3} x^{3/2}$$ - Apply formula: $$\int u dv = uv - \int v du$$ - So: $$\frac{2}{3} x^{3/2} \ln x - \int \frac{2}{3} x^{3/2} \cdot \frac{1}{x} dx = \frac{2}{3} x^{3/2} \ln x - \frac{2}{3} \int x^{1/2} dx$$ - Integrate $$\int x^{1/2} dx = \frac{2}{3} x^{3/2}$$ - Final answer: $$\frac{2}{3} x^{3/2} \ln x - \frac{2}{3} \cdot \frac{2}{3} x^{3/2} + C = \frac{2}{3} x^{3/2} \ln x - \frac{4}{9} x^{3/2} + C$$ 4. **Problem IV:** Evaluate $$\int \frac{dx}{\sqrt{x^2 - 6x + 10}}$$ - Complete the square inside the root: $$x^2 - 6x + 10 = (x^2 - 6x + 9) + 1 = (x - 3)^2 + 1$$ - Substitute $$u = x - 3$$, so $$du = dx$$ - Integral becomes: $$\int \frac{du}{\sqrt{u^2 + 1}}$$ - This is a standard form: $$\sinh^{-1}(u) + C = \ln|u + \sqrt{u^2 + 1}| + C$$ - Back-substitute: $$\boxed{\ln|x - 3 + \sqrt{x^2 - 6x + 10}| + C}$$ 5. **Problem V:** Evaluate $$\int \frac{3x^2 - 10}{x^2 - 4x + 4} dx$$ - Note denominator is $$ (x - 2)^2 $$ - Divide numerator by denominator: - Polynomial division: $$\frac{3x^2 - 10}{(x - 2)^2} = 3 + \frac{-12x + 2}{(x - 2)^2}$$ - So integral is: $$\int 3 dx + \int \frac{-12x + 2}{(x - 2)^2} dx = 3x + \int \frac{-12x + 2}{(x - 2)^2} dx$$ - Substitute $$u = x - 2$$, $$x = u + 2$$, $$dx = du$$ - Numerator: $$-12x + 2 = -12(u + 2) + 2 = -12u - 24 + 2 = -12u - 22$$ - Integral becomes: $$3x + \int \frac{-12u - 22}{u^2} du = 3x + \int \left(-12 \frac{u}{u^2} - 22 \frac{1}{u^2} \right) du = 3x + \int (-12 u^{-1} - 22 u^{-2}) du$$ - Integrate: $$3x - 12 \ln|u| + 22 u^{-1} + C = 3x - 12 \ln|x - 2| + \frac{22}{x - 2} + C$$ **Final answers:** I. $$\frac{\sec^{-1} \left( \frac{x}{\sqrt{\pi}} \right)}{\sqrt{\pi}} + C$$ II. $$e^t \left( x - \frac{1}{2} \ln(1 + e^{2x}) \right) + C$$ III. $$\frac{2}{3} x^{3/2} \ln x - \frac{4}{9} x^{3/2} + C$$ IV. $$\ln|x - 3 + \sqrt{x^2 - 6x + 10}| + C$$ V. $$3x - 12 \ln|x - 2| + \frac{22}{x - 2} + C$$