1. **Exercise 1a:** Calculate $I_a = \int \frac{dx}{x+3}$. Using the basic integral formula $\int \frac{dx}{x+c} = \ln|x+c| + C$, we get $$I_a = \ln|x+3| + C.$$ 2. **Exercise 1b:** Calculate $I_b = \int \frac{1-3x}{3+2x} dx$. Rewrite numerator: $1-3x = A(3+2x) + B$ to use substitution. Set $1-3x = A(3+2x) + B$. Equate coefficients: For $x$: $-3 = 2A \Rightarrow A = -\frac{3}{2}$. For constant: $1 = 3A + B = 3(-\frac{3}{2}) + B = -\frac{9}{2} + B \Rightarrow B = 1 + \frac{9}{2} = \frac{11}{2}$. So, $$I_b = \int \frac{-\frac{3}{2}(3+2x) + \frac{11}{2}}{3+2x} dx = \int \left(-\frac{3}{2} + \frac{11}{2(3+2x)}\right) dx.$$ Integrate termwise: $$I_b = -\frac{3}{2}x + \frac{11}{2} \int \frac{dx}{3+2x} = -\frac{3}{2}x + \frac{11}{2} \cdot \frac{1}{2} \ln|3+2x| + C = -\frac{3}{2}x + \frac{11}{4} \ln|3+2x| + C.$$ 3. **Exercise 1c:** Calculate $I_c = \int \frac{dx}{\sqrt{a^2 - x^2}}$. This is a standard integral: $$I_c = \arcsin\left(\frac{x}{a}\right) + C.$$ 4. **Exercise 1d:** Calculate $I_d = \int \frac{dx}{a^2 + x^2}$. Standard integral: $$I_d = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C.$$ 5. **Exercise 1e:** Calculate $I_e = \int \frac{dx}{\sqrt{x^2 + a^2}}$. Standard integral: $$I_e = \ln\left|x + \sqrt{x^2 + a^2}\right| + C.$$ 6. **Exercise 1f:** Calculate $I_f = \int \frac{dx}{x^2 - a^2}$. Use partial fractions: $$\frac{1}{x^2 - a^2} = \frac{1}{2a} \left(\frac{1}{x - a} - \frac{1}{x + a}\right).$$ Integrate: $$I_f = \frac{1}{2a} \ln\left|\frac{x - a}{x + a}\right| + C.$$ 7. **Exercise 1g:** Calculate $I_g = \int \frac{\cosh^3 x}{1 + \sinh x} dx$. Rewrite numerator: $\cosh^3 x = \cosh x \cdot \cosh^2 x = \cosh x (1 + \sinh^2 x)$. Set $t = \sinh x$, then $dt = \cosh x dx$. Rewrite integral: $$I_g = \int \frac{\cosh^3 x}{1 + \sinh x} dx = \int \frac{\cosh x (1 + t^2)}{1 + t} dx = \int \frac{1 + t^2}{1 + t} dt.$$ Divide numerator by denominator: $$\frac{1 + t^2}{1 + t} = t - 1 + \frac{2}{1 + t}.$$ Integrate: $$I_g = \int (t - 1) dt + 2 \int \frac{dt}{1 + t} = \frac{t^2}{2} - t + 2 \ln|1 + t| + C.$$ Back-substitute $t = \sinh x$: $$I_g = \frac{\sinh^2 x}{2} - \sinh x + 2 \ln|1 + \sinh x| + C.$$ 8. **Exercise 1h:** Calculate $I_h = \int \frac{\sqrt{x^2 - 4}}{x^4} dx$. Rewrite integrand: $$\frac{\sqrt{x^2 - 4}}{x^4} = \frac{\sqrt{x^2 - 4}}{x^4} = \frac{\sqrt{x^2 - 4}}{x^4}.$$ Use substitution $x = 2 \sec \theta$, then $dx = 2 \sec \theta \tan \theta d\theta$. Calculate: $$\sqrt{x^2 - 4} = \sqrt{4 \sec^2 \theta - 4} = 2 \tan \theta.$$ Rewrite integral: $$I_h = \int \frac{2 \tan \theta}{(2 \sec \theta)^4} \cdot 2 \sec \theta \tan \theta d\theta = \int \frac{2 \tan \theta}{16 \sec^4 \theta} \cdot 2 \sec \theta \tan \theta d\theta = \int \frac{4 \tan^2 \theta \sec \theta}{16 \sec^4 \theta} d\theta = \int \frac{\tan^2 \theta}{4 \sec^3 \theta} d\theta.$$ Rewrite $\sec^3 \theta = \frac{1}{\cos^3 \theta}$ and $\tan^2 \theta = \sec^2 \theta - 1$. So, $$I_h = \frac{1}{4} \int (\sec^2 \theta - 1) \cos^3 \theta d\theta = \frac{1}{4} \int (\cos \theta - \cos^3 \theta) d\theta.$$ Integrate: $$\int \cos \theta d\theta = \sin \theta, \quad \int \cos^3 \theta d\theta = \int \cos \theta (1 - \sin^2 \theta) d\theta = \sin \theta - \frac{\sin^3 \theta}{3} + C.$$ Therefore, $$I_h = \frac{1}{4} \left(\sin \theta - \sin \theta + \frac{\sin^3 \theta}{3}\right) + C = \frac{\sin^3 \theta}{12} + C.$$ Back-substitute $\sin \theta = \frac{\sqrt{x^2 - 4}}{x}$: $$I_h = \frac{1}{12} \left(\frac{\sqrt{x^2 - 4}}{x}\right)^3 + C = \frac{(x^2 - 4)^{3/2}}{12 x^3} + C.$$ **Final answers:** $$I_a = \ln|x+3| + C,$$ $$I_b = -\frac{3}{2}x + \frac{11}{4} \ln|3+2x| + C,$$ $$I_c = \arcsin\left(\frac{x}{a}\right) + C,$$ $$I_d = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C,$$ $$I_e = \ln\left|x + \sqrt{x^2 + a^2}\right| + C,$$ $$I_f = \frac{1}{2a} \ln\left|\frac{x - a}{x + a}\right| + C,$$ $$I_g = \frac{\sinh^2 x}{2} - \sinh x + 2 \ln|1 + \sinh x| + C,$$ $$I_h = \frac{(x^2 - 4)^{3/2}}{12 x^3} + C.$$