How do you integrate #{x(sqrt(1+x^2)} dx# from 0 to 1?

1 Answer
Ratnaker Mehta
Sep 22, 2016

#I=1/3(2sqrt2-1)#.

Explanation:

Let #I=int_0^1 {xsqrt(1+x^2)}dx.#

We subst. #1+x^2=t^2 rArr 2xdx=2tdt, i.e., xdx=tdt#

Further, #x=0 rArr t=1, and, x=1 rArr t=sqrt2#.

#:. I=int_1^(sqrt2) t.tdt = int_1^(sqrt2) t^2dt=[t^3/3]_1^sqrt2#.

#=1/3[sqrt2^3-1]#.

#:. I=1/3(2sqrt2-1)#.

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