Evaluate the integral. \int_{0}^{2}x\sqrt{x^{2}+1}dx

Trent Carpenter

Trent Carpenter

Answered question

2021-10-11

Evaluate the integral.
02xx2+1dx

Answer & Explanation

SchepperJ

SchepperJ

Skilled2021-10-12Added 96 answers

Step 1
To evaluate the given integrals.
Step 2
02xx2+1dx
Let x2+1=t
dtdx=2xdt2=xdx
Also At x =0 we have t =1
At x = 2 we have t = 5
Hence we have
02xx2+1dx=15tdt
=[t12+112+1]15
=23[t32]15=23[(5)32(1)32]
02xx2+1dx=23[(5)321]

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