\(I=\int\dfrac{dx}{\sqrt{\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}}}\)
Đặt \(x+\dfrac{3}{2}=\dfrac{1}{2}tanu\Rightarrow dx=\dfrac{1}{2cos^2u}du\)
\(I=\int\dfrac{1}{\dfrac{1}{2}\sqrt{tan^2u+1}}.\dfrac{1}{2.cos^2u}du=\int\dfrac{1}{cosu}du=\int\dfrac{1}{1-sin^2u}d\left(sinu\right)\)
\(=\dfrac{1}{2}ln\left|\dfrac{1+sinu}{1-sinu}\right|+C=ln\left(\dfrac{1+sinu}{cosu}\right)+C=ln\left(\dfrac{1}{cosu}+tanu\right)+C\)
Chú ý: \(\dfrac{1}{cosu}=\sqrt{\dfrac{1}{cos^2u}}=\sqrt{1+tan^2u}=\sqrt{1+\left(2x+3\right)^2}=2\sqrt{x^2+3x+2}\)
Do đó: \(I=ln\left(2x+3+2\sqrt{x^2+3x+2}\right)+C\)
Ủa giờ mới để ý tách biểu thức sai, \(x^2+3x+2=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\) mới đúng
Vậy làm cách khác:
Đặt \(\sqrt{\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}}=-\left(x+\dfrac{3}{2}\right)+t\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-2t\left(x+\dfrac{3}{2}\right)+t^2\)
\(\Rightarrow x+\dfrac{3}{2}=\dfrac{\dfrac{1}{4}+t^2}{2t}=\dfrac{1}{8t}+\dfrac{t}{2}\)
\(\Rightarrow dx=\left(-\dfrac{1}{8t^2}+\dfrac{1}{2}\right)dt=\left(\dfrac{4t^2-1}{8t^2}\right)dt\)
Lại có: \(\sqrt{x^2+3x+2}=-\left(x+\dfrac{3}{2}\right)+t=-\dfrac{1}{8t}-\dfrac{t}{2}+t=\dfrac{t}{2}-\dfrac{1}{8t}=\dfrac{4t^2-1}{8t}\)
\(\Rightarrow\dfrac{1}{\sqrt{x^2+3x+2}}=\dfrac{8t}{4t^2-1}\)
Do đó:
\(I=\int\dfrac{8t}{4t^2-1}.\dfrac{4t^2-1}{8t^2}=\int\dfrac{1}{t}dt=ln\left|t\right|+C\)
\(=ln\left|\sqrt{x^2+3x+2}+\left(x+\dfrac{3}{2}\right)\right|+C=ln\left|2\sqrt{x^2+3x+2}+2x+3\right|+C\)
Lần này chắc ko nhầm nữa :D
