integral x^2*sqrt(4x-1)dx

[jwpaine]

\(\displaystyle \int \,\, x^2 \sqrt{4x-1}\)

Substution doesn't get me anywhere..

\(\displaystyle u = 4x - 1, \,\,\frac{1}{4}du = dx, \,\,\frac{u + 1}{4} = x\)

\(\displaystyle \frac{1}{4}\int (\frac{u + 1}{4})^2 \sqrt{u} \,\,du\)

Which doesn't appear to get me anywhere.... do I have to construct partial fractions? We have only learned substitution method thus far.

 

[pka]

Just expand the expression.
\(\displaystyle \left( {u + 1} \right)^2 \sqrt u = u^{\frac{5}{2}} + 2u^{\frac{3}{2}} + u^{\frac{1}{2}}\)

 

[jwpaine]

Thanks.

 

[Deleted member 4993]

You could substitute

\(\displaystyle x = \frac{1}{4}\, sec^2(u)\)

\(\displaystyle dx = \frac{1}{2}\, sec(u)\cdot tan(u)\)