How do you evaluate #\int x ( x ^ { 2} + 1) ^ { 8} d x#?

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Answer 1 · 2017-11-07T08:55:23

#:.intx(x^2+1)^8dx=1/18(x^2+1)^9+C#

#intx(x^2+1)^8dx#

remember integration is the reverse operation of differentiation.

1) Inspection

we note that the function outside the bracket is the bracket differentiated, mutliplied by a constant.

so we suspect that the integral is the bracket tot eh power #8+1#

lets try it

#d/(dx)((x^+1)^9=9xx2x(x^2+1)^8=18x(x^2+1)^8#

which is the integrand except for the #18#

#:.intx(x^2+1)^8dx=1/18(x^2+1)^9+C#

#2) substitution

#intx(x^2+1)^8dx#

#u=x^2+1=>du=2xdx#

#intx(x^2+1)^8dx=intcancel(x)u^8xx(du)/(2cancel(x))#

#=1/2xx1/9u^9+c#

#=1/18u^9+C=1/18(x^2+1)^9+C#