2007-08-14 02:10:57 · 6 answers · asked by newbie4_z 1 in Science & Mathematics ➔ Mathematics
Let u = 2x + 1
du = 2 dx
du / 2 = dx
I = (1/2) ∫ 1 / (u) ³ du
I = (1/2) ∫ u^(- 3) + C
I = (- 1/4) u^(- 2) + C
I = (- 1) / [ (4)(2x + 1) ² ] + C
2007-08-14 02:51:24 · answer #1 · answered by Como 7 · 1⤊ 0⤋
I=1/2*int (2x+1)^(-3)d(2x+1)=
=-1/[4(2x+1)^2]+c
2007-08-14 09:49:25 · answer #2 · answered by ? 2 · 0⤊ 0⤋
Use a u substitution.
u = 2x + 1
du = 2dx
[1/(2x+1)^3]dx = 1/2 [1/u^3] du = 1/2 [u^-3] du
Integrate
= 1/2 [u^-2 / -2] + C
= -1/4 u^-2 + C
= -1/4 (2x+1)^-2 + C
= -1 / (4*(2x+1)^2 ) + C
2007-08-14 09:19:33 · answer #3 · answered by Jon G 4 · 1⤊ 0⤋
use the substitution method for Integral of 1/(2x+1)^3
Let u = 2x+1, then
du = 2dx or dx = du/2
integral = 1/u^3 du/2
= - 1/4u^2 + C
= - (1/4)1/(2x+1)^2 + C
2007-08-14 09:28:12 · answer #4 · answered by vlee1225 6 · 0⤊ 0⤋
integ (2x+1) ^-3
integ 1/2 (2x+1) ^-3 (2) = 1/2 (2x+1)^-2 /(-2) . . . power formula
= - 1/4 (2x+1) ^- 2 + c
= -1 / [ 4 (2x+1) ^2 ] + c
2007-08-14 09:27:40 · answer #5 · answered by CPUcate 6 · 0⤊ 0⤋
put 2x+1= t
differentiate on both side
2dx= dt
dx=1/2dt
Q.1/(2x+1)^3dx
substitute dx by dt
1/2(t)^3
integrating
-1/4(t)^2
= -1/4(2x+1)^2
2007-08-14 09:33:10 · answer #6 · answered by harsh_mkg 3 · 0⤊ 0⤋