Câu 1
Ta tính
$I = \displaystyle \int_1^2 \dfrac{x^2}{x^2 -7x +12}dx$
$= \displaystyle \int_1^2 \left( \dfrac{x^2-7x + 12}{x^2 - 7x + 12}+ \dfrac{7x-12}{x^2 - 7x + 12}\right)dx$
$= \displaystyle \int_1^2 \left( 1 + \dfrac{7x - 24,5}{x^2 - 7x + 12} + \dfrac{12,5}{(x-3)(x-4)} \right)dx$
$= \displaystyle \int_1^2 \left[ 1 + 3,5 \dfrac{2x-7}{x^2 - 7x + 12} + 12,5 \left( \dfrac{1}{x-4} - \dfrac{1}{x-3} \right) \right] dx$
$= \displaystyle \int_1^2 dx + 3,5 \displaystyle \int_1^2 \dfrac{2x-7}{x^2 - 7x + 12}dx + 12,5 \displaystyle \int_1^2 \left( \dfrac{1}{x-4} - \dfrac{1}{x-3} \right)dx$
$= \displaystyle \int_1^2 dx + 3,5 \displaystyle \int_1^2 \dfrac{d(x^2 - 7x + 12)}{x^2 - 7x + 12} + 12,5 \displaystyle \int_1^2 \left( \dfrac{1}{x-4} - \dfrac{1}{x-3} \right)dx$
$= \left[ x + 3,5 \ln|x^2 - 7x + 12| + 12,5(\ln|x-4| - \ln|x-3| ) \right]\Bigg\vert_1^2$
$= [2 + 3,5\ln 2 + 12,5 (\ln 2-0)] - [1 + 3,5 \ln 6 + 12,5(\ln 3 - \ln 2)]$
$= 25\ln 2 -16\ln 3 +1$
Câu 5
Ta tính
$I = \displaystyle \int_0^1 \dfrac{(x-1)^2}{(2x+1)^4}dx$
$= \displaystyle \int_0^1 \left( \dfrac{x-1}{2x+1} \right)^2 \dfrac{1}{(2x+1)^2}dx$
Ta có
$\left(\dfrac{x-1}{2x+1}\right)' = \dfrac{2x+1-2(x-1)}{(2x+1)^2} = \dfrac{3}{(2x+1)^2}$
Vậy
$d\left(\dfrac{x-1}{2x+1}\right) = \dfrac{3}{(2x+1)^2}dx$
$<-> \dfrac{1}{3} d\left(\dfrac{x-1}{2x+1}\right) = \dfrac{1}{(2x+1)^2}dx$
Thay vào ta có
$I = \dfrac{1}{3} \displaystyle \int_0^1 \left( \dfrac{x-1}{2x+1} \right)^2 d\left(\dfrac{x-1}{2x+1}\right)$
$= \dfrac{1}{3} \displaystyle \int_0^1 u^2 du$
$= \dfrac{1}{3} \dfrac{u^3}{3} \Bigg\vert_0^1$
$= \dfrac{1}{9}$
