How do you integrate #(x ^ { 2} + 8) ^ { 4} 2x d x#?

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Answer 1 · 2018-03-09T14:31:10

#int2x(x^2+8)^4dx=1/5(x^2+8)^5+c#

#int2x(x^2+8)^4dx#

one way is by inspection

we note that the outside of the bracket is a multiple of the bracket differentiated

so guess the bracket to the power #+1#

#d/(dx)((x^2+8)^5=2x xx 5(x^2+8)^4#

#=10x(x^2+8)^4#

comparing this with the integral we just adjust the multiple.

#int2x(x^2+8)^4dx=1/5(x^2+8)^5+c#