Question

Calcule a integral : ₁∫²(1/x + 1/x³)dx

Answer 1

Resposta:

de 1 a 2  ∫ (1/x + 1/x³)dx

∫  1/x dx =ln|x|

∫  1/x³  dx =x⁻²/(-2) =-1/2x²

de 1 a 2  [ ln|x| -1/2x²]

=ln|2| -1/8  -  [ln 1 -1/2]

= ln 2  - 1/8 - 0 +1/2

=ln 2 -1/8 +4/8

=ln 2 +3/8  

Answer 2

$\displaystyle\sf\int_{1}^{2}\bigg(\dfrac{1}{x} +\dfrac{1}{x^3}\bigg)~dx=\bigg[\ell n(x)-\dfrac{1}{2x^2}\bigg]_{1}^{2}\\\displaystyle\sf\int_1^2\bigg(\dfrac{1}{x}+\dfrac{1}{x^3}\bigg)~dx=\ell n(2)-\dfrac{1}{2\cdot2^2}-\bigg[\ell n(1)-\dfrac{1}{2\cdot1^2}\bigg] \\\displaystyle\sf\int_1^2\bigg(\dfrac{1}{x}-\dfrac{1}{x^3}\bigg)~dx=\ell n(2)-\dfrac{1}{8}-\ell n(1)+\dfrac{1}{2}\\\displaystyle\sf\int_1^2\bigg(\dfrac{1}{x}-\dfrac{1}{x^3}\bigg)~dx=\dfrac{8\ell n(2)-1+4}{8}$

$\large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int_1^2\bigg(\dfrac{1}{x}-\dfrac{1}{x^3}\bigg)~dx=\dfrac{8\ell n(2)+3}{8}}}}}$