How do you integrate #int10xsqrt(5x^2-4)# using substitution?

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Answer 1 · 2016-12-14T06:35:18

The answer is #=2/3(5x^2-4)^(3/2)+C#

We use, #intx^ndx=x^(n+1)/(n+1)+C##(n!=-1)#

Let #u=5x^2-4#, #=>##du=10xdx#

#int10xsqrt(5x^2-4)dx=intsqrtudu#

#=intu^(1/2)du=u^(3/2)/(3/2)#

#=2/3(5x^2-4)^(3/2)+C#

Answer 2 · 2016-12-16T06:20:03

See below, another way to solve this question:
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