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

2 Answers
Jun 26, 2017

#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#

Jun 26, 2017

# 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 #