Question

Evaluate `int \ (1+sqrt(x))^9/sqrt(x) \ dx`?

Answer 1

`int (sqrt(x)+1)^9/(sqrt(x)) "d"x = (sqrt(x)+1)^10/5 + C`

Explanation:

Could use the binomial theorem to expand `(1+sqrt(x))^9` and integrate simply from there but this seems tedious.

If we consider the integral,

`int "f"'(x)["f"(x)]^(n) "d"x`.

Substitute `u="f"(x)`. Then `"d"x = 1/("f"'(x)) "d"u`,

`int ("f"'(x)u^n)/("f"'(x)) "d"u` = (u*(n+1))/(n+1) + C#.

We conclude that,

`int "f"'(x)["f"(x)]^(n) "d"x = ["f"(x)]^(n+1)/(n+1) + C`.

This is a very common and very useful technique for solving integrals. Notice that `"d"/("d"x) sqrt(x) = 1/(2sqrt(x))`.

We can therefore rewrite the given integral easily so it is in this form.

`int (sqrt(x)+1)^9/(sqrt(x)) "d"x = 2 int 1/(2sqrt(x)) (sqrt(x)+1)^9 "d"x`.

Using this general result we conclude,

`int (sqrt(x)+1)^9/(sqrt(x)) "d"x = 2( (sqrt(x)+1)^10/10 ) + C`
`int (sqrt(x)+1)^9/(sqrt(x)) "d"x = (sqrt(x)+1)^10/5 + C`

Answer 2

`int \ (1+sqrt(x))^9/sqrt(x) \ dx = (1+sqrt(x))^10/5 + C`

Explanation:

We want to evaluate:

`I = int \ (1+sqrt(x))^9/sqrt(x) \ dx`

We can perform a simple substitution. Let:

`u = 1+sqrt(x) => (du)/dx = 1/(2sqrt(x))`

Substituting into the integral we get:

`I = int \ u^9 \ 2 \ du`
`\ \ = 2 \ int \ u^9 \ du`
`\ \ = 2 u^10/10 + C`
`\ \ = u^10/5 + C`

Restoring the substitution we get:

`I = (1+sqrt(x))^10/5 + C`