How do you find #\int ( 2x + \frac { 3} { x ^ { 2} } ) d x#?

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Answer 1 · 2017-02-05T21:36:06

#int(2x+3/x^2)dx=x^2-3/x+C#

Since #int f(x)+g(x)dx=intf(x)dx+intg(x)dx#

We can say

#int(2x+3/x^2)dx=int2xdx+int3/x^2dx#

Also since #intcxdx=cintxdx#

we can say

#int2xdx+3/x^2dx=2intxdx+3int1/x^2dx=2(x^2/2)-3(1/x)#

Then

#int(2x+3/x^2)dx=x^2-3/x+C#