How do you find #\int _ { 0} ^ { 2} \frac { x } { \sqrt { 2x ^ { 2} + 1} } d x#?

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Answer 1 · 2017-04-01T00:03:32

#int_0^2x/sqrt(2x^2+1)dx=1#

Use the substitution #u=2x^2+1#. This implies that #du=4xcolor(white).dx#.

When we switch from #x# to #u#, we will have to transform the bounds.

#x=0" "=>" "u=2(0^2)+1=1#

#x=2" "=>" "u=2(2^2)+1=9#

So:

#int_0^2x/sqrt(2x^2+1)dx=1/4int_0^2(4x)/sqrt(2x^2+1)dx=1/4int_1^9 1/sqrtudu#

Rewriting with exponents and integrating:

#=1/4int_1^9u^(-1/2)du=[1/4(u^(1/2)/(1/2))]_1^9=[1/2sqrtu]_1^9=1/2sqrt9-1/2sqrt1=1#