Question

Integral definida (método de substituição)
$\int\limits^1_0 \frac{1}{1+x} \, dx$

Answer 1

$\int\limits_{0}^{1}\dfrac{1}{1+x}dx$

Seja u = 1 + x, então du = dx

Para x = 0, u = 1
Para  x = 1, u = 2

Portanto:

$\int\limits_{0}^{1}\dfrac{1}{1+x}dx=\int\limits_{1}^{2}\dfrac{1}{u}du\\\\\\\int\limits_{0}^{1}\dfrac{1}{1+x}dx=ln(u)]_{1}^{2}\\\\\\\int\limits_{0}^{1}\dfrac{1}{1+x}dx=ln(2)-ln(1)\\\\\\\int\limits_{0}^{1}\dfrac{1}{1+x}dx=ln(2)-0\\\\\\\boxed{\boxed{\int\limits_{0}^{1}\dfrac{1}{1+x}dx=ln(2)}}$