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

Question

Charles Arocha · Accepted Answer

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

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#

Evelyn Agard · Answer

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

Elijah Abram · Answer

undefined The antiderivative of $ \frac{2x}{\sqrt{5 + 4x}} $ can be found by using a u-substitution. Let $ u = 5 + 4x $, then $ du = 4 dx $, or equivalently, $ \frac{1}{4} du = dx $.

Substitute $ u $ and $ du $ into the integral:
$$ \int \frac{2x}{\sqrt{5 + 4x}} dx = \int \frac{2}{\sqrt{u}} \cdot \frac{u - 5}{4} du $$
$$ = \frac{1}{2} \int u^{-\frac{1}{2}} u du - \frac{5}{2} \int u^{-\frac{1}{2}} du $$
$$ = \frac{1}{2} \left( \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right) - \frac{5}{2} \left( \frac{u^{\frac{-1}{2}}}{\frac{-1}{2}} \right) + C $$
$$ = u^{\frac{1}{2}} + 5u^{\frac{-1}{2}} + C $$
$$ = \sqrt{5 + 4x} + \frac{5}{\sqrt{5 + 4x}} + C $$

So, the antiderivative of $ \frac{2x}{\sqrt{5 + 4x}} $ is $ \sqrt{5 + 4x} + \frac{5}{\sqrt{5 + 4x}} + C $, where $ C $ is the constant of integration.