How do you evaluate the integral #int (xdx)/(5x^2-2)#?

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Answer 1 · 2018-02-05T18:28:34

# 1/10ln|(5x^2-2)|+C#.

Prerequisite : #int{(f'(x))/f(x)}dx=ln|f(x)|+c#.

This can be proved using subst. for #f(x)#.

Knowing taht, #d/dx(5x^2-2)=10x#, we have,

# int(xdx)/(5x^2-2)=1/10int(10x)/(5x^2-2)dx#,

#=1/10ln|(5x^2-2)|+C#.

Answer 2 · 2018-02-05T18:30:55

#intx/(5x^2-2)dx=1/10ln|(5x^2-2)|+c#

we use the result

#int(f'(x))/(f(x))dx=ln|f(x)|+c#

#intx/(5x^2-2)dx--(1)#

now #d/(dx)(5x^2-2)=10x#

rewriting #(1)#

#1/10int(10x)/(5x^2-2)dx#

we have

#intx/(5x^2-2)dx=1/10ln|(5x^2-2)|+c#