T_TEngineer_AdamT_T
New member
- Joined
- Apr 15, 2007
- Messages
- 24
\(\displaystyle \L \int \frac{dx}{\sqrt{2+\sqrt{x-1}}}\)
solutions:
let z^2 = x-1
2zdz = dx
\(\displaystyle \L \int \frac{2zdz}{\sqrt{2+z}}\)
i used integration buy parts:
let dv = \(\displaystyle \L \frac{dz}{\sqrt{2+z}}\)
v = \(\displaystyle \L 2\sqrt{2+z}\)
u = 2z
du = 2dz
\(\displaystyle \L 2z(2\sqrt{z+2}) - \int {4z\sqrt{z+2}}\)
\(\displaystyle \L 4z\sqrt{z+2} - 4{\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\)
\(\displaystyle \L 4(\sqrt{x-1})\sqrt{\sqrt{x-1}+2} - \frac{8}{3}(2+\sqrt{x-1})^{\frac{3}{2}} + C\)
the book's answer is :
\(\displaystyle \L \frac{4}{3}\sqrt{2+\sqrt{x-1}}(\sqrt{x-1} - 4) + C\)
solutions:
let z^2 = x-1
2zdz = dx
\(\displaystyle \L \int \frac{2zdz}{\sqrt{2+z}}\)
i used integration buy parts:
let dv = \(\displaystyle \L \frac{dz}{\sqrt{2+z}}\)
v = \(\displaystyle \L 2\sqrt{2+z}\)
u = 2z
du = 2dz
\(\displaystyle \L 2z(2\sqrt{z+2}) - \int {4z\sqrt{z+2}}\)
\(\displaystyle \L 4z\sqrt{z+2} - 4{\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\)
\(\displaystyle \L 4(\sqrt{x-1})\sqrt{\sqrt{x-1}+2} - \frac{8}{3}(2+\sqrt{x-1})^{\frac{3}{2}} + C\)
the book's answer is :
\(\displaystyle \L \frac{4}{3}\sqrt{2+\sqrt{x-1}}(\sqrt{x-1} - 4) + C\)
