How do you integrate #e^7x^3 x^2 dx#?

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Answer 1 · 2016-08-15T17:54:20

#=(e^7)/6 (x^6) +C #

#int e^7x^3x^2 \dx#

I can see two approaches 1) combining the #x# terms or 2) doing u substitution. first lets express this differently by moving out the constant

#e^7int x^3x^2 \dx#
1)

#e^7int x^5\\dx#
#(e^7)/6 (x^6)+C#

2)
now i will use u substitution and let #u=x^3# then
#(du)/(dx) =3x^2#
#1/3du=x^2 dx#

now substituting this back in
#e^7int u1/3 \\du#
#(e^7)/3int u \\du#

#=(e^7)/3 (u^2)/2 +C#
now we place replace u
#=(e^7)/6 (x^3)^2 +C #
#=(e^7)/6 (x^6) +C #