Cho hàm số f(x) có đạo hàm liên tục trên [0;1] thỏa mãn f(1)=0; $\int\limits^1_0 {[f'(x)]^{2} } \, dx$ =$\frac{3}{2}$-2ln2 và $\int\limits^1_0 {\frac{f(x)}{(x+1)^{2}}}\, dx$ =2ln2-$\frac{3}{2}$ . Tính tích phân $\int\limits^1_0 {f(x)} \, dx$

Đáp án: $ \displaystyle\int^1_0f(x)dx=-\dfrac{1}{2}\left(-\dfrac{1}{2}\ln ^2\left(2\right)-\ln \left(2\right)+1\right)$

Giải thích các bước giải:

Ta có:

$\displaystyle\int^1_0(f'(x))^2dx=\dfrac32-2\ln2$

$\displaystyle\int^1_0\dfrac{f(x)}{(x+1)^2}dx=2\ln2-\dfrac32$

$\to \displaystyle\int^1_0(f'(x))^2dx+\displaystyle\int^1_0\dfrac{f(x)}{(x+1)^2}dx=\dfrac32-2\ln2+2\ln2-\dfrac32$

$\to \displaystyle\int^1_0(f'(x))^2+\dfrac{f(x)}{(x+1)^2}dx=0$

$\to (f'(x))^2+\dfrac{f(x)}{(x+1)^2}=0$

$\to (f'(x))^2=-\dfrac{f(x)}{(x+1)^2}$

$\to \dfrac{(f'(x))^2}{-f(x)}=\dfrac1{(x+1)^2}$

$\to \dfrac{f'(x)}{\sqrt{-f(x)}}=\dfrac1{x+1}$

$\to\displaystyle\int \dfrac{f'(x)}{\sqrt{-f(x)}}dx=\displaystyle\int\dfrac1{x+1}dx$

$\to -2\displaystyle\int \dfrac{-f'(x)}{2\sqrt{-f(x)}}dx=\displaystyle\int\dfrac1{x+1}dx$

$\to -2\cdot \sqrt{-f(x)}=\ln|x+1|+ C$

Mà $f(1)=0\to -2\cdot \sqrt{-f(1)}=\ln|1+1|+ C$

$\to \ln2+C=0\to C=-\ln 2$

$\to -2\cdot \sqrt{-f(x)}=\ln|x+1|-\ln2$

$\to -2\cdot \sqrt{-f(x)}=\ln\dfrac{|x+1|}{2}$

$\to 4(-f(x))=\ln^2\dfrac{|x+1|}{2}$

$\to f(x)=-\dfrac14\ln^2\dfrac{|x+1|}{2}$

$\to \displaystyle\int^1_0f(x)dx=\displaystyle\int^1_0-\dfrac14\ln^2\dfrac{|x+1|}{2}dx$

$\to \displaystyle\int^1_0f(x)dx=\displaystyle\int^1_0-\dfrac14\ln^2\dfrac{x+1}{2}dx$

$\to \displaystyle\int^1_0f(x)dx=-\dfrac{1}{2}\left(-\dfrac{1}{2}\ln ^2\left(2\right)-\ln \left(2\right)+1\right)$