What will be the Evaluate the following integrals ? (1) ∫ (a/x^2 + b/x) dx (2) ∫ (3x^-7+x^-1) (3) ∫ (√x+1/√x)^2 dx

1 Answer
Steve M
Jun 16, 2018

(1) # int \ a/x^2+b/x \ dx = -a/x + bln|x| + c #

(2) # int \ 3x^(-7)+x^(-1) \ dx = -1/(2x^6) + ln|x| + c #

(3) # int \ (sqrt(x) +1/sqrt(x))^2 \ dx =x^2/2+2x + ln|x| + c #

Explanation:

We use the power rule:

# int \ x^n \ dx = x^(n+1)/(n+1) \ \ \ (+c) \ \ \ AA x in RR, x!= -1#

And the standard result:

# int \ 1/x \ dx = lnx \ \ \ \ (+c)#

So that:

Part (1):

# I_1 = int \ a/x^2+b/x \ dx #

# \ \ \ = int \ ax^(-2)+b/x \ dx #

# \ \ \ = ax^(-1)/(-1) + bln|x| + c #

# \ \ \ = -a/x + bln|x| + c #

Part (2):

# I_2 = int \ 3x^(-7)+x^(-1) \ dx #

# \ \ \ = 3x^(-6)/(-6) + ln|x| + c #

# \ \ \ = -1/(2x^6) + ln|x| + c #

Part (3):

# I_3 = int \ (sqrt(x) +1/sqrt(x))^2 \ dx #

# \ \ \ = int \ (sqrt(x))^2 + 2 sqrt(x)1/sqrt(x) + (1/sqrt(x))^2 \ dx #

# \ \ \ = int \ x + 2 + 1/x \ dx #

# \ \ \ = x^2/2+2x + ln|x| + c #

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