$I= \int\limits_{0}^{1}\frac{x^3-\sqrt{1-x}}{x+3}dx=\int\limits_{0}^{1}\frac{x^3}{x+3}dx-\int\limits_{0}^{1}\frac{\sqrt{1-x}}{x+3}dx=I_1-I_2$
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: $I=I_1-I_2=\frac{59}{6}-27ln\frac{4}{3}-2ln3.$