Integral:#int(12x^3)/sqrt(2x^2+7)dx# ?

#int(12x^3)/sqrt(2x^2+7)dx#

2 Answers
Apr 11, 2018

#sqrt(2x^2+7) (2x^2-14)+C#

Explanation:

Substitute #2x^2+7=u^2# so that #4xdx = 2udu# and #x^2=(u^2-7)/2#

Thus

#int(12x^3)/sqrt(2x^2+7)dx=int(6x^2)/sqrt(2x^2+7) 2xdx#
#qquad = int {3(u^2-7)}/{u} udu = 3int (u^2-7)du#
#qquad = u^3-21u+C = u(u^2-21)+C#
#qquad =sqrt(2x^2+7) (2x^2-14)+C#

Apr 11, 2018

#I=2sqrt(2x^2+7)(x^2-7)+c#

Explanation:

Here,

#I=int(12x^3)/sqrt(2x^2+7)dx#

#=int(12x^2)/sqrt(2x^2+7)*xdx#

Let, #2x^2+7=u^2=>2x^2=u^2-7=>4xdx=2udu#

#=>xdx=1/2udu#

So,

#I=int(12(u^2-7)/2)/canceluxx1/2canceludu#

#=3int(u^2-7)du#

#=3[u^3/3-7u]+c#

#=u^3-21u+c...to# where , #2x^2+7=u^2#

#=u(u^2-21)+c#

#=sqrt(2x^2+7)(2x^2+7-21)+c#

#=sqrt(2x^2+7)(2x^2-14)+c#

#=2sqrt(2x^2+7)(x^2-7)+c#