Question

What is the antiderivative of `(2x)/(sqrt(5 + 4x))`?

Answer 1

`sqrt(5+4x)/6(2x-5)+c`

Explanation:

To find the antiderivertive of `(2x)/(sqrt(5 + 4x))` we calculate the integral

`int((2x)/(sqrt(5 + 4x)))dx`

substitute `u=5+4x` and `du=4dx`
`int((u-5)/(8sqrt(u)))du=1/8(int(u/sqrt(u))du-int(5/sqrt(u))du)`

`int(5/sqrt(u))du=10sqrt(u)=10sqrt(5+4x)`
`int(u/sqrt(u))du=int(sqrt(u))du=2/3sqrt(u^3)=2/3sqrt((5+4x)^3)`
All in all:

`sqrt(5+4x)/6(2x-5)+c`

Answer 2

`int(2x)/sqrt(5+4x)dx=1/6(2x-5)sqrt(5+4x)+C`

Explanation:

We are asked to find

`int(2x)/sqrt(5+4x)dx`

Let's first substitute

`u=5+4x`

`rArrdu=4dx`

`1/4du=dx`

`rArru-5=4x`

`rArr1/2(u-5)=2x`

Now the integral becomes:

`int(1/2(u-5))/sqrtu1/4du`

`rArr1/8int(u-5)/sqrtudu`

Let's split this into two simpler integrals:

`rArr1/8int(u/sqrtu-5/sqrtu)du`

`rArr1/8[intsqrtudu-5int1/sqrtudu]`

... and let's express the roots as exponents to make integrating more intuitive.

`rArr1/8[intu^(1/2)du-5intu^(-1/2)du]`

Integrating, we get:

`rArr1/8[u^(3/2)/(3/2)-5u^(1/2)/(1/2)]+C`

`rArr1/8[2/3u^(3/2)-10u^(1/2)]+C`

`rArr2/24u^(3/2)-10/8u^(1/2)+C`

`rArr1/12u^(3/2)-5/4u^(1/2)+C`

`rArr1/4u^(1/2)(1/3u-5)+C`

Now let's express the answer in terms of `x`, and perform some algebra to simplify the expression.

`rArr1/4sqrt(5+4x)(1/3(5+4x)-5)+C`

`rArr1/4sqrt(5+4x)(1/3(5+4x-15))+C`

`rArr1/4sqrt(5+4x)(1/3(4x-10))+C`

`rArr1/12(4x-10)sqrt(5+4x)+C`

`rArr1/12*2(2x-5)sqrt(5+4x)+C`

`rArr1/6(2x-5)sqrt(5+4x)+C`