Tính tích phân $I=\int _2^3 \sqrt{x^2+10x-23}\mbox{d}x$

Tính tích phân $I=\displaystyle \int _{2}^{3} \sqrt{x^2+10x-23} \mbox{d}x$

Dạng $\displaystyle \int \sqrt{ax^2+bx+c} \mbox{d}x$
Đặt

$\begin{matrix}u=\sqrt{x^2+10x-23} &\Rightarrow& du=\dfrac{x+5}{\sqrt{x^2+10x-23}}dx \\ dv=dx &\Rightarrow& v=x \end{matrix}$

$\begin{align}I &= x\sqrt{x^2+10x-23}\bigg|_{2}^{3}-\displaystyle \int _{2}^{3} \dfrac{x^2+5x}{\sqrt{x^2+10x-23}}\mbox{d}x \\ &= 10-\displaystyle \int _{2}^{3} \dfrac{(x^2+10x-23)-5(x+5)+48}{\sqrt{x^2+10x-23}}\mbox{d}x \\ &= 10-I+5\displaystyle \int _{2}^{3} \dfrac{x+5}{\sqrt{x^2+10x-23}}\mbox{d}x -48\displaystyle \int _{2}^{3} \dfrac{1}{\sqrt{x^2+10x-23}}\mbox{d}x \end{align}$

$2I = 10+5I_1-48I_2$

Tính $I_1$

Đặt $t=\sqrt{x^2+10x-23} \Rightarrow t.dt=(x+5)dx \\  x=2 \Rightarrow t=1 , x=3 \Rightarrow t=4$

$I_1=\displaystyle \int _{1}^{4} \mbox{d}t=3$

Tính $I_2$

Đặt $t_2=\sqrt{x^2+10x-23} +x+5 \Rightarrow dt_2=\dfrac{\sqrt{x^2+10x-23} +x+5}{\sqrt{x^2+10x-23}}dx $

$\Rightarrow \dfrac{1}{t_2}\mbox{d}t_2=\dfrac{1}{\sqrt{x^2+10x-23}}\mbox{d}x$

$ x=2 \Rightarrow t_2=7 , x=3 \Rightarrow t_2=12$

$I_2=\displaystyle \int _{7}^{12} \dfrac{1}{t_2}\mbox{d}t_2=\ln|t_2| \bigg|_{7}^{12}=\ln \dfrac{12}{7}$

Vậy $I= \dfrac{25}{2}-24\ln \dfrac{12}{7}$