Question

vol=$\pi \int\limits^1_0 { \frac{1}{( x+1)^{2} } } \, dx$

Answer 1

( x + 1 )  '  / (x +1 ) ² = ( x +1 ) '  · ( x +1 ) ⁻ ² = u ' ·u ⁻²  integrat 
 = - ( x +1 ) ⁻ ¹ = - 1 / ( x +1 ) 
Vol = π [  - 1 / 2 + 1/1  ] = π /2

Answer 2

[tex]u(x)=x+1, u'(x)dx=1dx=du, u(0)=1,u(1)=2\\ Vol=\pi \int\limits^1_0 {\frac{1}{(x+1)^2}(x+1)'} \, dx=\pi \int\limits^1_0 {u^{-2}(x)u'(x)} \, dx= \\ \pi \int\limits^2_1 {u^{-2}} \, du=\pi(\frac{u^{-1}}{-1})|_1^2=-\pi(\frac{1}{2}-\frac{1}{1})=\frac{\pi}{2} [/tex]