Question

Evaluate the integral

∫_0^1{36dx/((2x+1)^3)}

Answer 1

`36 \int\limits^1_0 { (1)/((2 x + 1) ^(3) ) } \, dx =`
u = 2 x + 1, d u = 2 d x, d x = d u / 2

`36*1/2 \int\limits^1_0 {u ^(-3) } \, du = 18 * t ^(-2)/(-2) = \\ - 9 * t^(-2)= \\ -9 * ( 2 x + 1 ) ^(-2)`
- 9 / ( 2 + 1 )² - ( - 9 / ( 0 + 1 )² =
- 9 / 9 + 9 = - 1 + 9 = 8