Como se resuelven parte 2 integrales

$$\begin{align}& \end{align}$$

¡Hola Myriam!

·

$$\begin{align}&e)\quad \int_0^3 \left(\frac{x^3}{2}-2x^2+x+3 \right)dx=\\&\\&\left[\frac{x^4}{8}-\frac{2x^3}{3}+\frac {x^2}{2}+3x  \right]_0^3=\\&\\&\frac{81}{8}-18+\frac{9}{2}+9-0+0-0-0=\\&\\&\frac{81+36-72}{8}= \frac{45}{8}\\&\\&\\&\\&f)  \int_2^6 \frac{x}{\sqrt{5x^2+1}}dx=\\&\\&t=5x^2+1\\&dt = 10x\, dx\implies x\,dx=\frac 1{10}dt\\&x=2\implies t=5·2^2+1=21\\&x=6\implies t=5·6^2+1=181\\&\\&=\frac 1{10}\int_{21}^{181}\frac{dt}{ \sqrt t} =\frac 1{10}\int_{21}^{181}t^{-1/2}dt =\\&\\&\frac 1{10}\left. \frac{t^{1/2}}{\frac 12} \right|_{21}^{181}=\left.\frac 1{5} t^{1/2} \right|_{21}^{181}= \\&\\&\\&\frac{\sqrt{181}-\sqrt{21}}{5}\\&\\&\\&\end{align}$$

Y eso es todo.

e) Esta se hace directa, sin sustitución:

$$\begin{align}&\int()dx=\\&\\&\frac{1}{2} ·\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+3x]_2^6=\\&\\&\frac{81}{8}-\frac{54}{3}+\frac{9}{2}+9-0=\frac{45}{8}\\&\\&f) Sustitución:\\&5x^2+1=t\\&10xdx=dt\\&\int \frac{x}{\sqrt{5x^2+1}}dx= \int \frac{dt}{10 \sqrt t}=\frac{1}{10} \int t^{-\frac{1}{2}}dt=\\&\\&\frac{1}{10} \frac {t^{\frac{-1}{2}+1}}{1+\frac{-1}{2}}=\frac{1}{5}t^{\frac{1}{2}}= \frac{\sqrt {t}}{5}=\\&\\&\frac{1}{5} \sqrt{5x^2+1}\\&\\&\left[ \frac{1}{5} \sqrt{5x^2+1} \right]_2^6=\frac{1}{5}(\sqrt{81}-\sqrt{21})\end{align}$$