Integrate ? #3sqrt(4x+3) dx #

2 Answers
Apr 3, 2018

#int3sqrt(4x+3)dx=((4x+3)sqrt(4x+3))/2+C#

Explanation:

.

#int3sqrt(4x+3)dx#

Let #u=4x+3, :. du=4dx, :.dx=(du)/4#

Le's substitute:

#int3sqrtu(du)/4=3/4intu^(1/2)du=3/4(2/3)u^(3/2)=1/2u^(3/2)#

Let's substitute back:

#int3sqrt(4x+3)dx=1/2(4x+3)^(3/2)=1/2sqrt((4x+3)^3)#

#int3sqrt(4x+3)dx=((4x+3)sqrt(4x+3))/2+C#

Apr 3, 2018

#3intsqrt(4x+3) dx #

Let #(4x+3) = t#
#=> 4 dx = dt #

#3intsqrt(4x+3) dx = 3int sqrtt dt /4#

#=> 3/4int sqrtt dt#

#=> 3/4 (t^(3/2)) /(3/2) +c #

#=> 3/4 xx 2/3 (t^(3/2)) +c#

#=> 1/2 (4x+3)^(3/2) +c#