Cho biết nguyên hàm từ 1 đến 2 ln(9−?2)??=?ln5+?ln2+? với a,b,c là các số nguyên. Tính ?=?+?+?

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} u=\ln \left(9-x^2\right)\Rightarrow du=\frac{-2x}{9-x^2}dx \\ dv=dx\Rightarrow v=x \end{gathered}$$

$$\begin{gathered} \underset{1}{\overset{2}{\int }}ln\left(9-x^2\right)dx={\left.x.ln\left(9-x^2\right)\right|}_1^2-\underset{1}{\overset{2}{\int }}\frac{-2x^2}{9-x^2}dx \\ ={\left.x.ln\left(9-x^2\right)\right|}_1^2+\underset{1}{\overset{2}{\int }}\frac{2x^2}{9-x^2}dx \\ ={\left.x.ln\left(9-x^2\right)\right|}_1^2+\underset{1}{\overset{2}{\int }}\left(-2+\frac{{\displaystyle 18}}{{\displaystyle 9-x^2}}\right)dx \end{gathered}$$

$$\begin{gathered} ={\left.x.\ln \left(9-x^2\right)\right|}_1^2-{\left.2x\right|}_1^2+\underset{1}{\overset{2}{\int }}3\left(\frac{1}{3-x}+\frac{1}{3+x}\right)dx \\ ={\left.x.\ln \left(9-x^2\right)\right|}_1^2-{\left.2x\right|}_1^2+{\left.3\ln \left|\frac{3+x}{3-x}\right|\right|}_1^2 \\ =5\ln 5-6\ln 2-2. \end{gathered}$$

Vậy $S=5-6-2=-3.$