Tính tích phân sau: I=∫1+sin x+x.sinx/ (x+1)dx

Question

Tính tích phân sau:

\(I=\int _{0}^{\pi }\frac{1+\sin x+x\sin x}{x+1} {\rm d}x\)

Answer 1

\(I=\int _{0}^{\pi }\frac{1+\sin x+x\sin x}{x+1} {\rm d}x =\int _{0}^{\pi }\frac{1+\left(1+x\right)\sin x}{x+1} {\rm d}x \)

\(=\int _{0}^{\pi }\left(\frac{1}{x+1} +\sin x\right){\rm d}x  \)
\(=\int _{0}^{\pi }\frac{1}{x+1} {\rm d}x+ \int _{0}^{\pi }\sin x{\rm d}x\)

\(=\left. \ln \left|x+1\right|\right|_{0}^{\pi } -\left. \cos x\right|_{0}^{\pi } =\ln \left(\pi +1\right)+2. \)