Question

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

Answer 1

`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`

Answer 2

`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`