Resposta:

1- Integre a derivada para encontrar a f(x).

[tex]\sf f'(x)=1+ln(x)[/tex]

[tex]\sf f(x)=\int\sf(1+ln(x))dx[/tex]

[tex]\sf f(x)=\int\sf1dx+\int\sf ln(x)dx[/tex]

[tex]\sf f(x)=x+\int\sf ln(x)dx[/tex]

Integrando por partes, denote respectivamente, por u e dv:

• [tex]\sf u=ln(x)\implies du=\frac{1}{x}dx[/tex]

• [tex]\sf dv=dx\implies v=x[/tex]

[tex]\sf f(x)=x+u\cdot v-\int\sf v\cdot du[/tex]

[tex]\sf f(x)=x+ln(x)\cdot x-\int\sf x\cdot\frac{1}{x}dx[/tex]

[tex]\sf f(x)=x+ln(x)\cdot x-\int\sf dx[/tex]

[tex]\sf f(x)=x+ln(x)\cdot x-x+\mathnormal{C}[/tex]

[tex]\sf f(x)=ln(x)\cdot x+\mathnormal{C}[/tex]

Dado que f(1) = 1:

[tex]\sf f(1)=ln(1)\cdot 1+\mathnormal{C}=1[/tex]

[tex]\sf ln(1)\cdot 1+\mathnormal{C}=1[/tex]

[tex]\sf 0+\mathnormal{C}=1[/tex]

[tex]\sf \mathnormal{C}=1[/tex]

Portanto:

[tex]\sf f(x)=ln(x)\cdot x+1[/tex]

Letra D

2- Calcule a integral indefinida primeiro e aplique o teorema fundamental do cálculo depois pra calcular a integral definida.

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int^{\sf e}_{\sf1}\sf\big(\frac{1}{x}+\sqrt{x^5}\big)dx[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf\big(\frac{1}{x}+\sqrt{x^5}dx\big)\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf\frac{1}{x}dx+\int\sf\sqrt{x^5}dx\,\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf\frac{1}{x}dx+\int\sf (x^5)^{\frac{1}{2}}dx\,\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\int\sf \frac{1}{x}dx+\int\sf x^{\frac{5}{2}}dx\,\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|x|+\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}\,\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|x|+\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\,\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|x|+\frac{2x^{\frac{7}{2}}}{7}\,\big|^{\sf e}_{\sf1}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=ln|e|+\frac{2(e)^{\frac{7}{2}}}{7}-(ln|1|+\frac{2(1)^{\frac{7}{2}}}{7})[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=1+\frac{2e^{\frac{7}{2}}}{7}-(0+\frac{2(1)}{7})[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=1+\frac{2\sqrt{e^7}}{7}-\frac{2}{7}[/tex]

[tex]\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\frac{7\,+\,2\sqrt{e^7}\,-\,2}{7}[/tex]

[tex]\red{\sf \int^{\sf e}_{\sf1}\sf f(x)\,dx=\frac{5\,+\,2\sqrt{e^7}}{7}}[/tex]

Letra E