Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

∫^[infinity]_3 1/(x+1)^3 dx

Answer 1

Convergent;
`(1)/(32)`.

Explanation:

We have been given an integral as
`\int _3^(\infty )\:\:(1)/(\left(x+1\right)^3)\:dx`. We are asked to determine whether our given integral converges or diverges.

Let us integrate our given integral by u substitution as:

`\int _3^(\infty )\:\:(1)/(\left(u)^3)\:dx`

`\int _3^(\infty )\:\:u^(-3)\:dx`

`\int _3^(\infty )\:\:u^(-3)\:dx=(u^(-3+1))/(-3+1)`

`\int _3^(\infty )\:\:u^(-3)\:dx=(u^(-2))/(-3+1)`

`((x+1)^(-2))/(-2)=-(1)/(2(x+1)^2)`

Now, we will compute the boundaries.

`-(1)/(2(\infty+1)^2)=-(1)/(\infty ^2)=0`

`-(1)/(2(3+1)^2)=-(1)/(2(4^2)=-(1)/(2*16)=-(1)/(32)`

Our definite integral would be
`0-(-(1)/(32))=(1)/(32)`

Therefore, our given integral is convergent and its value is
`(1)/(32)`.

Answer 2

`I=(1)/(32)`

Explanation:

Given,

Improper Integral I is given as

`I=\int^(\infty)_(3)(1)/((x+1)^3)dx`

integration of

`(1)/((x+1)^2)`

now

`\int (1)/((x+1)^3)dx = ((x+1)^(-3+1))/(-3+1)`

`\int (1)/((x+1)^3)dx =(1)/(-2(x+1)^2)`

`I=\left [(1)/(-2(x+1)^2)\right]^(\infty)_3`

substituting value

`I=-(1)/(2)\left [ (1)/(\infty^2)-(1)/(4^2)\right ]`

`I=-(1)/(2)\left [-(1)/(16)\right ]`

`I=(1)/(32)`

so the value of integral converges at
`(1)/(32)`