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

Question

2870 views around the world

You can reuse this answer
Creative Commons License

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

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

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