Cho hàm số \(y=f\left( x \right)\) thỏa mãn điều kiện \(\int\limits_{0}^{2}{\frac{f'\left( x \right)dx}{x+2}}=3\) và \(f\left( 2 \right)-2f\left( 0 \right)=4\). Tính tích phân \(I=...

* Hướng dẫn giải

Đặt \(\left\{ \begin{align} & u=\frac{1}{x+2} \\ & dv=f'\left( x \right)dx \\ \end{align} \right.\)

\(\Rightarrow \left\{ \begin{align} & du=-\frac{1}{{{\left( x+2 \right)}^{2}}} \\ & v=f\left( x \right) \\ \end{align} \right.\)

Khi đó \(\int\limits_{0}^{2}{\frac{f'\left( x \right)dx}{x+2}}=\frac{f\left( x \right)}{x+2}\left| _{0}^{2} \right.+\int\limits_{0}^{2}{\frac{f\left( x \right)dx}{{{\left( x+2 \right)}^{2}}}}=\frac{f\left( 2 \right)}{4}-\frac{f\left( 0 \right)}{2}+\int\limits_{0}^{2}{\frac{f\left( x \right)dx}{{{\left( x+2 \right)}^{2}}}}=1+\int\limits_{0}^{2}{\frac{f\left( x \right)dx}{{{\left( x+2 \right)}^{2}}}}\).

Suy ra \(K=\int\limits_{0}^{2}{\frac{f\left( x \right)dx}{{{\left( x+2 \right)}^{2}}}}=2\xrightarrow{x=2t}K=\int\limits_{0}^{1}{\frac{f\left( 2t \right)d2t}{{{\left( 2t+2 \right)}^{2}}}}=\int\limits_{0}^{1}{\frac{f\left( 2t \right)dt}{2{{\left( t+1 \right)}^{2}}}}=2\).

Vậy \(\int\limits_{0}^{1}{\frac{f\left( 2t \right)dt}{{{\left( t+1 \right)}^{2}}}}=4\).