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#