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).

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 u12 u12 du2 equal 14 inte class=

Respuesta :

[tex]\bf \displaystyle \int x\sqrt{2x-1}\cdot dx\\\\ -----------------------------\\\\ u=2x-1\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\ -----------------------------\\\\ \displaystyle \int x\sqrt{u}\cdot \cfrac{du}{2}\\\\ -----------------------------\\\\ u=2x-1\implies u+1=2x\implies \cfrac{u+1}{2}=x\\\\ -----------------------------\\\\[/tex]

[tex]\bf \displaystyle \int \cfrac{u+1}{2}\sqrt{u}\cdot \cfrac{du}{2}\implies \int \left(\cfrac{u+1}{2} \right)u^{\frac{1}{2}}\cdot \cfrac{du}{2} \\\\\\ \displaystyle\int \cfrac{u^{\frac{3}{2}+}u^{\frac{1}{2}}}{2}\cdot \cfrac{du}{2}\implies \cfrac{1}{4}\int u^{\frac{3}{2}}\cdot du+\cfrac{1}{4}\int u^{\frac{1}{2}}\cdot du[/tex]

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