Đáp án:
$37)\quad A.\,I=\dfrac12\displaystyle\int\limits_1^2\dfrac{(t-1)^3}{t^5}dt$
$38)\quad A. \, I = \dfrac{1502.2^{1001}}{501501}$
Giải thích các bước giải:
$37)\quad I =\displaystyle\int\limits_0^1\dfrac{x^7}{(1+x^2)^5}dx$
Đặt $t = 1 + x^2\longrightarrow x^2 = t -1$
$\to dt = 2xdx$
Đổi cận:
$x\quad \Big|\quad 0\qquad 1$
$\overline{\, t\quad \Big|\quad 1\qquad 2}$
Ta được:
$\quad I = \dfrac12\displaystyle\int\limits_0^1\dfrac{(x^2)^3}{(1+x^2)^5}2xdx$
$\to I =\dfrac12\displaystyle\int\limits_1^2\dfrac{(t-1)^3}{t^5}dt$
$38)\quad I =\displaystyle\int\limits_1^3x(x-1)^{1000}dx$
Đặt $u= x-1\longrightarrow x = u+1$
$\to du= dx$
Đổi cận:
$x\quad \Big|\quad 1\qquad 3$
$\overline{u\quad \Big|\quad 0\qquad 2}$
Ta được:
$\quad I =\displaystyle\int\limits_0^2(u+1)u^{1000}du$
$\to I = \displaystyle\int\limits_0^2u^{1001}du + \displaystyle\int\limits_0^2u^{1000}du$
$\to I =\dfrac{u^{1002}}{1002}\Bigg|_0^2 + \dfrac{u^{1001}}{1001}\Bigg|_0^2$
$\to I =\dfrac{2^{1002}}{1002} +\dfrac{2^{1001}}{1001}$
$\to I =\dfrac{1001.2^{1002} + 1002.2^{1001}}{1003002}$
$\to I = \dfrac{2^{1001}(1001.2+1002)}{1003002}$
$\to I =\dfrac{2^{1001}.3004}{1003002}$
$\to I =\dfrac{2^{1001}.1502}{501501}$
