How do you evaluate the integral #int x/sqrt(4x+1)#?

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Answer 1 · 2017-04-10T20:19:25

#int x/sqrt(4x+1) dx = (2x-1)/12sqrt(4x+1) +C#

Note that:

#1/sqrt(4x+1) = 1/2 d/dx sqrt(4x+1)#

So we can write the integral as:

#int x/sqrt(4x+1) dx = 1/2 int x d(sqrt(4x+1))#

and integrate by parts:

#int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/2 int sqrt(4x+1)dx#

The resulting integral can be resolved directly using the power rule:

#int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/8 int (4x+1)^(1/2)d(4x+1)#

#int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/8 (4x+1)^(3/2)/(3/2)+C#

and simplifying:

#int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/12 (4x+1)sqrt(4x+1)+C#

#int x/sqrt(4x+1) dx = (1/2x-1/3x-1/12)sqrt(4x+1) +C#

#int x/sqrt(4x+1) dx = (2x-1)/12sqrt(4x+1) +C#