How do you integrate #\int _ { 2} ^ { - 3} ( 3+ x ^ { 2} ) ^ { 2} d x #?

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Answer 1 · 2017-11-22T12:31:37

The answer is #=-170#

We need

#intx^ndx=x^(n+1)/(n+1)+ C (x!=-1)#

The integral is

#int_2^-3(3+x^2)^2dx= int_2^-3(9+6x^2+x^4)dx#

#=[9x+2x^3+x^5/5]_2^-3#

#=(-27-54-243/5)-(18+16+32/5)#

#=-81-34-275/5#

#=-170#