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Hola Luis!
$$\begin{align}&\int _0^1(2x-3)dx= x^2-3x \Bigg |_0^1=(1-3)-0=-2\\&\\&\int_1^2 \frac{5-x}{x^3} dx= \int 5x^{-3}-x^{-2}dx= 5 \frac{x^{-2}}{-2}- \frac{x^{-1}}{-1}=\frac{-5}{2}x^{-2}+x^{-1} \Bigg |_1^2=\\&\\&\frac{-5}{2} 2^{-2}+2^{-1}-\Big (\frac{-5}{2}+1^{-1}\Big)= \frac{-5}{8}+ \frac{1}{2}+\frac{5}{2}-1= \frac{11}{8}\\&\\&\int_1^5 2 (x-1)^{\frac{1}{2}}=2 \frac{(x-1)^{\frac{3}{2}}}{\frac{3}{2}}= \frac{4}{3} \Bigg [(x-1)^\frac{3}{2} \Bigg]_1^5= \frac{4}{3} \Big(4^\frac{3}{2}-0 \Big)= \frac{4}{3}·8=\frac{32}{3}\end{align}$$
saludos
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