How do you integrate #int xsqrt(2x^2+7)# using substitution?

1 Answer
Gió
Jan 22, 2017

I got: #intxsqrt(2x^2+7)dx=1/6(2x^2+7)sqrt(2x^2+7)+c#

Explanation:

Let us set:
#2x^2+7=t#
derive:
#4x=dt#
let us use this inside our integral (red):
#int1/4color(red)(4x)sqrt(t)color(red)(dx)=#
where I multiplyed and divided by #4#:
substituting the red part with the derived bit we get:
#int1/4sqrt(t)dt=#
integrating:
#1/4intt^(1/2)dt=1/4t^(1/2+1)/(1/2+1)+c=1/4t^(3/2)/(3/2)+c=#
#=1/6tsqrt(t)+c#

but: #2x^2+7=t#

so we get:
#intxsqrt(2x^2+7)dx=1/6(2x^2+7)sqrt(2x^2+7)+c#

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