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Question

$\int\limits_{0}^{1}\frac{x^3-\sqrt{1-x}}{x+3}dx$

★★.P.I.N.O.★★ · Answer 1 · 6/17/2014 2:33:22 PM

$I= \int\limits_{0}^{1}\frac{x^3-\sqrt{1-x}}{x+3}dx=\int\limits_{0}^{1}\frac{x^3}{x+3}dx-\int\limits_{0}^{1}\frac{\sqrt{1-x}}{x+3}dx=I_1-I_2$

Tính

$I_1.$

Đặt

$x+3=t$ , ta có:

$dx=dt$

Đổi cận:

$x=0\rightarrow t=3, x=1\rightarrow t=4$

$I_1=\int\limits_{3}^{4}\frac{(t-3)^3}{t}dt=\int\limits_{3}^{4}(t^2-9t+27-\frac{27}{t})dt=(\frac{t^3}{3}-\frac{9t^2}{2}+27t-27lnt)\bigg|_3^4$

$=\frac{47}{6}-27ln\frac{4}{3}.$

Tính

$I_2.$

Đặt

$\sqrt{1-x}=a$ , ta có:

$1-x=a^2\Rightarrow -dx=2ada$

Đổi cận:

$x=0\rightarrow a=1, x=1\rightarrow a=0.$

$I_2=\int\limits_{1}^{0}\frac{2a^2}{a^2-4}da=2\int\limits_{1}^{0}(1+\frac{1}{a-2}-\frac{1}{a+2})da=2(a+ln\left| {\frac{a-2}{a+2}} \right|)\bigg|_1^0$

$=-2+2ln3$

KQ:

$I=I_1-I_2=\frac{59}{6}-27ln\frac{4}{3}-2ln3.$

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