Karekök x'in İntegrali

Karekök x'in İntegrali Nedir ?

Karekök x'in (√x) integrali (2/3).x.√x+c'dir.

$\int \sqrt{x}dx=\frac{2}{3}.x.\sqrt{x}+c$

1. Yol

$\sqrt{x}=x^{\frac{1}{2}}$

$\int \sqrt{x}dx=\int x^{\frac{1}{2}}dx$

$\boxed{\int x^{n}dx=\frac{x^{n+1}}{n+1}+c}$

$\int \sqrt{x}dx=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$\int \sqrt{x}dx=\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c$

$\int \sqrt{x}dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c$

$\int \sqrt{x}dx=\frac{2}{3}.x^{\frac{3}{2}}+c$

$\int \sqrt{x}dx=\frac{2}{3}.\sqrt{x^3}+c$

$\int \sqrt{x}dx=\frac{2}{3}.\sqrt{x^2.x}+c$

$\int \sqrt{x}dx=\frac{2}{3}.x.\sqrt{x}+c$

2. Yol

$\int \sqrt{x}dx=?$

$\sqrt{x}=u$

$(\sqrt{x}{)}^2=u^2$

$x=u^2$

$dx=d(u^2)$

$dx=(u^2{)}^{\prime }du$

$dx=2.udu$

$\int \sqrt{x}dx=\int u.2.udu$

$\int \sqrt{x}dx=\int 2.u^{2}du$

$\int \sqrt{x}dx=\frac{2.u^{2+1}}{2+1}+c$

$\int \sqrt{x}dx=\frac{2.u^3}{3}+c$

$\int \sqrt{x}dx=\frac{2}{3}.u^3+c$

$\int \sqrt{x}dx=\frac{2}{3}.(\sqrt{x}{)}^3+c$

$\int \sqrt{x}dx=\frac{2}{3}.\sqrt{x^3}+c$

$\int \sqrt{x}dx=\frac{2}{3}.\sqrt{x^2.x}+c$

$\int \sqrt{x}dx=\frac{2}{3}.x.\sqrt{x}+c$