How do you integrate by substitution #int tsqrt(t^2+2)dt#?

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Answer 1 · 2017-01-17T13:56:03

The answer is #=(t^2+2)^(3/2)/3+C#

Let #u^2=t^2+2#

Differentiating both sides

#2udu=2tdt#, #=>#, #tdt=udu#

Therefore,

#inttsqrt(t^2+2)dt=intu*udu#

#intu^2du=u^3/3#

#=(t^2+2)^(3/2)/3+C#