Tính tích phân $I=\int_1^2 \frac{4x+3}{\sqrt{x^2+5x-5}} dx$

Tính tích phân $I=\displaystyle \int_1^2 \dfrac{4x+3}{\sqrt{x^2+5x-5}} dx$



$ \begin{matrix} I &=& \displaystyle \int_1^2 \dfrac{2(2x+5)-7}{\sqrt{x^2+5x-5}}dx \\ &=& \displaystyle 2\int_1^2 \dfrac{2x+5}{\sqrt{x^2+5x-5}}dx -7 \displaystyle \int_1^2 \dfrac{1}{\sqrt{x^2+5x-5}}dx &=& 2I_1 -7I_2 \end{matrix} $

Tính $I_1$
Đặt $t=\sqrt{x^2+5x-5} \Rightarrow 2t.dt=(2x+5)dx$
$x=1 \Rightarrow t=1 , \,\,  x=2 \Rightarrow t=3$

$I_1= \displaystyle \int_1^3 2dt =2t \bigg|_{1}^{3}=4$

Tính $I_2$
$ \begin{matrix} t_2 &=& 2\sqrt{x^2+5x-5}+2x+5 \\ \Rightarrow dt_2 &=& \left( \dfrac{2x+5}{\sqrt{x^2+5x-5}}+2 \right) dx = \dfrac{2\sqrt{x^2+5x-5}+2x+5}{\sqrt{x^2+5x-5}}dx \\ \Rightarrow \dfrac{1}{t_2}dt_2 &=&\dfrac{1}{\sqrt{x^2+5x-5}} dx \end{matrix}$

$x=1 \Rightarrow t_2=9 , \,\,  x=2 \Rightarrow t_2=15$

$I_2= \displaystyle \int_9^{15} \dfrac{1}{t_2}dt_2=\ln |t_2| \bigg|_{9}^{15}= \ln \dfrac{5}{3}$

Ta có : $I = 2I_1 -7I_2 =8-7\ln \dfrac{5}{3}$