Question

Determine a primitiva da integral:

A. −[ln(x)+1]x+c

B. ln(x)−1x+c

C. ln2(x)+c

D. ln(x)+c

E. −ln(x)+1+c

Answer 1

Resposta:  $\displaystyle\int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=-\,\frac{\mathrm{\ell n}(x)+1}{x}+C.$

Explicação passo a passo:

Calcular a integral indefinida:

    $\displaystyle\int \frac{\mathrm{\ell n}(x)}{x^2}\,dx\\\\\\=\int \mathrm{\ell n}(x)\cdot x^{-2}\,dx$

Integrando por partes:

    $\begin{array}{lcl}u=\mathrm{\ell n}(x)&\quad\Longrightarrow\quad&du=\dfrac{1}{x}\,dx=x^{-1}\,dx\\\\dv=x^{-2}\,dx&\quad\Longleftarrow\quad&v=-\,x^{-1} \end{array}$

    $\displaystyle\int u\,dv=uv-\int v\,du\\\\\\\Longrightarrow\quad \int \mathrm{\ell n}(x)\cdot x^{-2}\,dx=\mathrm{\ell n}(x)\cdot (-\,x^{-1})-\int(-\,x^{-1})\cdot x^{-1}\,dx\\\\\\\Longleftrightarrow\quad \int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=-\,\frac{\mathrm{\ell n}(x)}{x}+\int x^{-2}\,dx\\\\\\\Longleftrightarrow\quad \int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=-\,\frac{\mathrm{\ell n}(x)}{x}+\frac{x^{-2+1}}{-\,2+1}+C$

    $\displaystyle\Longleftrightarrow\quad \int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=-\,\frac{\mathrm{\ell n}(x)}{x}+\frac{x^{-1}}{-1}+C\\\\\\\Longleftrightarrow\quad \int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=-\,\frac{\mathrm{\ell n}(x)}{x}-\frac{1}{x}+C\\\\\\\Longleftrightarrow\quad \int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=\frac{-\,\mathrm{\ell n}(x)-1}{x}+C\\\\\\\Longleftrightarrow\quad \int \frac{\mathrm{\ell n}(x)}{x^2}\,dx=-\,\frac{\mathrm{\ell n}(x)+1}{x}+C\quad\longleftarrow\quad\mathsf{resposta.}$

Dúvidas? Comente.

Bons estudos!