Question

I do not understand this integration problem. I understand why they made u, du, and dx what they did, no problems. But Isnt the integral of (u+1)/2 * u^(1/2) du/2

equal 1/4 integral u * u^(1/2)? I do not understand where the second line comes from with the u^(3/2)+u^(1/2).

Answer 1

`\bf \displaystyle \int x√(2x-1)\cdot dx\\\\ -----------------------------\\\\ u=2x-1\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\ -----------------------------\\\\ \displaystyle \int x√(u)\cdot \cfrac{du}{2}\\\\ -----------------------------\\\\ u=2x-1\implies u+1=2x\implies \cfrac{u+1}{2}=x\\\\ -----------------------------\\\\`


`\bf \displaystyle \int \cfrac{u+1}{2}√(u)\cdot \cfrac{du}{2}\implies \int \left(\cfrac{u+1}{2} \right)u^{(1)/(2)}\cdot \cfrac{du}{2} \\\\\\ \displaystyle\int \cfrac{u^{(3)/(2)+}u^{(1)/(2)}}{2}\cdot \cfrac{du}{2}\implies \cfrac{1}{4}\int u^{(3)/(2)}\cdot du+\cfrac{1}{4}\int u^{(1)/(2)}\cdot du`


`\bf \cfrac{1}{4}\cdot \cfrac{u^{(5)/(2)}}{(5)/(2)}+\cfrac{1}{4}\cdot \cfrac{u^{(3)/(2)}}{(3)/(2)}\implies \cfrac{1}{4}\cdot \cfrac{2u^{(5)/(2)}}{5}+\cfrac{1}{4}\cdot \cfrac{2u^{(3)/(2)}}{3}\implies \cfrac{u^{(5)/(2)}}{10}+\cfrac{u^{(3)/(2)}}{6} \\\\\\ \cfrac{1}{10}(2x-1)^{(5)/(2)}+\cfrac{1}{6}(2x-1)^{(3)/(2)}`