find integral of 1/(2x+7)^2?

Answer 1

I = ∫ 1 / (2x + 7) ² dx
let u = 2x + 7
du = 2 dx
dx = du / 2
I = (1/2) ∫ u ^ (-2) du
I = (- 1 / 2) u ^ (-1) + C
I = - 1 / [ 2 (2x + 7) ] + C

Answer 2

Are you familiar with u-substitution?

U-sub is the concept that if you can do the chain rule:

d[g( f(x) )]/dx = g'( f(x)) * f'(x)

That this can go into the opposite direction. Integrate both sides:

g( f(x)) = integral[ g'( f(x)) * f'(x) ]

Therefore, if you can find out f'(x) --or what is re-defined as U' , where U is the inside-- you can integrate normally by just getting rid of it.

Therefore, U in this case is what's inside: 2x+7. dU or U' = 2, correct?

So how do you shift this? You multiply the top and the bottom by 2, take the two you need on the top, and then leave the remaining two on the bottom. Then integrate normally:

int[ 1/2(U)^-2 ] = -1/4 U

U = 2x+7

Sub it back in, and that's your answer. Ask your teacher again about switching between differentials (dx => dU), because that'll help you a lot.

Answer 3

Set y = 2x + 7; then dy/dx = 2, or dx = (1/2) dy

integral ( 1/y^2 dx)
= integral y^(-2) (1/2) dy
= (1/2) integral y^-2 dy
= (1/2) (-1/3) y^-3 + C
= - (1/6) y^-3 + C
= - (1/6) (2x+7)^-3 + C

Answer 4

integral (2x+7)^-2. put into u du form:
(2x+7)^-2 * 2 *1/2, and so -(2x+7)^-1 * 1/2 + c, or
-1/2(2x+7) + c

Answer 5

substitute u = 2x + 7 and you will be able to find an antiderivative in u.