Answer
$0.19080$
Work Step by Step
We need to solve the integral. $I=\int_{0}^{0.2} \dfrac{\ln (1+t)}{t} \ dt$
Now, we have:
$I=\int_{0}^{0.2} \dfrac{\ln (1+t)}{t} \ dt \\=\int_{0}^{0.2}(\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{t^{k-1}}{k} ) \ dt\\=|\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{t^{k}}{k^2}|_{0}^{0.2}\\=\Sigma_{k =1}^{\infty} (-1)^{k+1} \dfrac{0.2^k}{k^2}$
By using a calculator, we get:
$I=0.19080$
