\(I_2=\int\sqrt{\left(x^2+2x+1\right)+4}dx=\int\sqrt{\left(x+1\right)^2+4}dx\)
Đặt \(x+1=2tant\Rightarrow dx=\dfrac{2}{cos^2t}dt\)
\(I_2=\int\sqrt{4tan^2t+4}.\dfrac{2}{cos^2t}dt=\int\dfrac{2}{cost}.\dfrac{2}{cos^2t}dt\)
Xét \(I=\int\dfrac{1}{cos^3t}dt\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{cost}\\dv=\dfrac{dt}{cos^2t}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{sint}{cos^2t}dt\\v=tant=\dfrac{sint}{cost}\end{matrix}\right.\)
\(\Rightarrow I_2=\dfrac{sin}{cos^2t}-\int\dfrac{sin^2t}{cos^3t}dt=\dfrac{sint}{cos^2t}-\int\dfrac{1-cos^2t}{cos^3t}dt\)
\(=\dfrac{sint}{cos^3t}-\int\dfrac{1}{cos^3t}dt+\int\dfrac{1}{cost}dt=\dfrac{sint}{cos^3t}-I+\int\dfrac{1}{cost}dt\) (1)
Xét \(\int\dfrac{1}{cost}dt=\int\dfrac{1}{cost}.\dfrac{tant+\dfrac{1}{cost}}{tant+\dfrac{1}{cost}}dt=\int\dfrac{\dfrac{1}{cos^2t}+\dfrac{sint}{cos^2t}}{tant+\dfrac{1}{cost}}dt\)
\(=\int\dfrac{d\left(tant+\dfrac{1}{cost}\right)}{tant+\dfrac{1}{cost}}=ln\left|tant+\dfrac{1}{cost}\right|+C\) (2)
(1);(2) \(\Rightarrow I=\dfrac{sint}{cos^2t}-I+ln\left|tant+\dfrac{1}{cost}\right|+C\)
\(\Rightarrow2I=\dfrac{sint}{cos^2t}+ln\left|tant+\dfrac{1}{cost}\right|+C\)
\(\Rightarrow I_2=4I=\dfrac{2sint}{cos^2t}+2ln\left|tant+\dfrac{1}{cost}\right|+C\)
Trả biến:
\(x+1=2tant\Rightarrow\dfrac{\left(x+1\right)^2}{4}=tan^2t\)
\(\Rightarrow\dfrac{\left(x+1\right)^2}{4}+1=1+tan^2t=\dfrac{1}{cos^2t}\)
\(\Rightarrow\dfrac{1}{cost}=\sqrt{\dfrac{\left(x+1\right)^2}{4}+1}\)
\(\dfrac{sint}{cos^2}=tant.\dfrac{1}{cost}=\dfrac{\left(x+1\right)}{2}.\sqrt{\dfrac{\left(x+1\right)^2}{4}+1}\)
Do đó:
\(I_2=\left(x+1\right)\sqrt{\dfrac{\left(x+1\right)^2}{4}+1}+2ln\left|\dfrac{x+1}{2}+\sqrt{\dfrac{\left(x+1\right)^2}{4}+1}\right|+C\)
Cách 2: sử dụng phép thế Euler có vẻ nhẹ nhàng hơn:
Đặt \(\sqrt{x^2+2x+5}=t-x\) (1)
\(\Rightarrow x^2+2x+5=t^2-2tx+x^2\)
\(\Rightarrow x\left(2t+2\right)=t^2-5\)
\(\Rightarrow x=\dfrac{t^2-5}{2t+2}\)
\(\Rightarrow dx=\dfrac{t^2+2t+5}{2\left(t+1\right)^2}dt\)
Đồng thời (1) \(\Rightarrow\sqrt{x^2+2x+5}=t-x=t-\dfrac{t^2-5}{2t+2}=\dfrac{t^2+2t+5}{2t+2}\)
\(\Rightarrow I_2=\int\dfrac{t^2+2t+5}{2t+2}.\dfrac{t^2+2t+5}{2\left(t+1\right)^2}dt=\dfrac{1}{4}\int\dfrac{\left[\left(t+1\right)^2+4\right]^2}{\left(t+1\right)^3}dt\)
\(=\dfrac{1}{4}\int\dfrac{\left(t+1\right)^4+8\left(t+1\right)^2+16}{\left(t+1\right)^3}dt\)
\(=\dfrac{1}{4}\int\left(t+1+\dfrac{8}{t+1}+\dfrac{16}{\left(t+1\right)^3}\right)dt\)
\(=\dfrac{\left(t+1\right)^2}{8}+2ln\left|t+1\right|-\dfrac{2}{\left(t+1\right)^2}+C\)
\(=\dfrac{\left(x+1+\sqrt{x^2+2x+5}\right)^2}{8}+2ln\left|x+1+\sqrt{x^2+2x+5}\right|-\dfrac{2}{\left(x+1+\sqrt{x^2+2x+5}\right)^2}+C\)
\(=\dfrac{\left(x+1+\sqrt{x^2+2x+5}\right)^2}{8}-\dfrac{2\left(x+1-\sqrt{x^2+2x+5}\right)^2}{16}+2ln\left|x+1+\sqrt{x^2+2x+5}\right|+C\)
\(=\dfrac{\left(x+1\right)\sqrt{x^2+2x+5}}{2}-2ln\left|x+1+\sqrt{x^2+2x+5}\right|+C\)
