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

1 Answer
Mar 9, 2018

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

Explanation:

#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#