Question

\(\int\limits^1_0xln\left(2+x^2\right)dx=aln3+bln2+c\) tính a+b+c

Answer 1

Đặt \(\left\{{}\begin{matrix}u=ln\left(2+x^2\right)\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{2x}{x^2+2}dx\\v=\frac{x^2}{2}\end{matrix}\right.\)

\(\Rightarrow I=\frac{x^2}{2}.ln\left(2+x^2\right)|^1_0-\int\limits^1_0\frac{x^3}{x^2+2}dx=\frac{1}{2}ln3-I_1\)

\(I_1=\int\limits^1_0\frac{x^3}{x^2+2}dx=\int\limits^1_0\left(x-\frac{2x}{x^2+2}\right)dx=\left(\frac{x^2}{2}-ln\left(x^2+2\right)\right)|^1_0=\frac{1}{2}-ln3+ln2\)

\(\Rightarrow I=\frac{3}{2}ln3-ln2-\frac{1}{2}\)\(\Rightarrow a+b+c=\frac{3}{2}-1-\frac{1}{2}=0\)