Küpkök x'in İntegrali

Küpkök x'in İntegrali Nedir ?

Küpkök x'in (∛x) integrali (3/4).x.∛x+c'dir.

$\int \sqrt[3]{x}dx=\frac{3}{4}.x.\sqrt[3]{x}+c$

1. Yol

$\sqrt[3]{x}=x^{\frac{1}{3}}$

$\int \sqrt[3]{x}dx=\int x^{\frac{1}{3}}dx$

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

$\int \sqrt[3]{x}dx=\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}+c$

$\int \sqrt[3]{x}dx=\frac{x^{\frac{1+3}{3}}}{\frac{1+3}{3}}+c$

$\int \sqrt[3]{x}dx=\frac{x^{\frac{4}{3}}}{\frac{4}{3}}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.x^{\frac{4}{3}}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.\sqrt[3]{x^4}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.\sqrt[3]{x^3.x}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.x.\sqrt[3]{x}+c$

2. Yol

$\int \sqrt[3]{x}dx=?$

$\sqrt[3]{x}=u$

$(\sqrt[3]{x}{)}^3=u^3$

$x=u^3$

$dx=d(u^3)$

$dx=(u^3{)}^{\prime }du$

$dx=3.u^{2}du$

$\int \sqrt[3]{x}dx=\int u.3.u^{2}du$

$\int \sqrt[3]{x}dx=\int 3.u^{3}du$

$\int \sqrt[3]{x}dx=\frac{3.u^{3+1}}{3+1}+c$

$\int \sqrt[3]{x}dx=\frac{3.u^4}{4}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.u^4+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.(\sqrt[3]{x}{)}^4+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.\sqrt[3]{x^4}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.\sqrt[3]{x^3.x}+c$

$\int \sqrt[3]{x}dx=\frac{3}{4}.x.\sqrt[3]{x}+c$