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Question

$\int\limits_{0}^{1}x^{3}.\ln (\frac{4-x^{2}}{4+x^{2}})dx$

score 1 · Answer 1

$I=\int_{0}^{1} x^3.\ln (\dfrac{4-x^2}{4+x^2}) =\dfrac{1}{2}\int_{0}^{1} x^2\ln\dfrac{4-x^2}{4+x^2}d{x^2}=\dfrac{1}{2} \int_{0}^{1} t\ln\dfrac{4-t}{4+t}dt$

Đặt $\begin{cases} u=\ln\dfrac{4-t}{4+t} \\ dv = tdt \end{cases} \implies \begin{cases} du=\dfrac{8}{t^2-16}dt\\ v=\dfrac{t^2}{2} \end{cases}$

Khi đó $ I=\bigg (\dfrac{t^2}{4}\ln\dfrac{4-t}{4+t}\bigg ) \bigg |_{0}^{1}- 2\int_{0}^{1} \dfrac{t^2}{t^2-16}dt$

$ =\dfrac{1}{4} \ln \dfrac{3}{5}- 2 \int_{0}^{1} (1+\dfrac{16}{t^2-16})dt=\dfrac{1}{4} \ln \dfrac{3}{5}- \bigg (2t +4 \ln \dfrac{t-4}{t+4} \bigg) \bigg|_{0}^{1} =...$