[tex]I=\int \dfrac{x+2}{\sqrt{x^2+3x+2}}=\int \dfrac{2x+3}{2\sqrt{x^2+3x+2}}dx+\dfrac{1}{2}\int \dfrac{1}{\sqrt{x^2+3x+2}}dx=\sqrt{x^2+3x+2}+\dfrac{1}{2}I_1[/tex]
Xét [tex]I_1[/tex], đặt [tex]x+\dfrac{3}{2}=\dfrac{1}{2cost}>0\Rightarrow dx=\dfrac{sint}{2cos^2t}dt[/tex]
[tex]I_1=\int \dfrac{1}{\sqrt{\dfrac{1}{4cos^2t}-\dfrac{1}{4}}}.\dfrac{sint}{2cos^2t}dt=\int \dfrac{1}{cost}dt=\int \dfrac{d(sint)}{1-sin^2t}=\dfrac{1}{2}ln\left | \dfrac{1+sint}{1-sint} \right |+C[/tex]
[tex]=\dfrac{1}{2}ln\left | \dfrac{(1+sint)^2}{cos^2t} \right |+C=ln\left | \dfrac{1+sint}{cost} \right |+C=ln\left | \dfrac{1}{cost}+tant \right |+C[/tex]
[tex]=ln\left | 2x+3+2\sqrt{x^2+3x+2} \right |+C[/tex]
[tex]\Rightarrow I=\sqrt{x^2+3x+2}+\dfrac{1}{2}ln\left | 2x+3+2\sqrt{x^2+3x+2} \right |+C[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
