Question

Tính nguyên hàm \(\int\dfrac{1}{x^3+x}dx\)

Answer 1

\(\int\dfrac{dx}{x^3+x}=\int\dfrac{dx}{x\left(x^2+1\right)}\)

\(t=x^2+1\Rightarrow dt=2xdx\Rightarrow\int\dfrac{dx}{x\left(x^2+1\right)}=\int\dfrac{dt}{2x^2t}=\dfrac{1}{2}\int\dfrac{dt}{\left(t-1\right).t}\)

\(\dfrac{1}{\left(t-1\right).t}=\dfrac{1}{t-1}-\dfrac{1}{t}\)

\(\Rightarrow\int\dfrac{dt}{\left(t-1\right)t}=\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\int\dfrac{dt}{t-1}-\int\dfrac{dt}{t}=ln\left|t-1\right|-ln\left|t\right|=ln\left|x^2\right|-ln\left|x^2+1\right|\)