What is the integral of (x-2)/(\sqrt[3]{x^2-4x+6)} ?

The integral of (x-2)/(\sqrt[3]{x^2-4x+6)} is 3/4 (x^2-4x+6)^{2/3}+C

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integral of (x-2)/(\sqrt[3]{x^2-4x+6)}

\int \frac{x - 2}{3 x ^{2} - 4 x + 6} d x

\frac{3}{4} (x^{2} - 4 x + 6)^{\frac{2}{3}} + C

Solution steps

\int \frac{x - 2}{3 x ^{2} - 4 x + 6} d x

Apply u-substitution

: \int \frac{1}{2 3 u} d u

= \int \frac{1}{2 3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{3 u} d u

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{2} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = x^{2} - 4 x + 6

= \frac{1}{2} \cdot \frac{3}{2} (x^{2} - 4 x + 6)^{\frac{2}{3}}

Simplify

\frac{1}{2} \cdot \frac{3}{2} (x^{2} - 4 x + 6)^{\frac{2}{3}} : \frac{3}{4} (x^{2} - 4 x + 6)^{\frac{2}{3}}

= \frac{3}{4} (x^{2} - 4 x + 6)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{4} (x^{2} - 4 x + 6)^{\frac{2}{3}} + C

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What is the integral of (x-2)/(\sqrt[3]{x^2-4x+6)} ?
The integral of (x-2)/(\sqrt[3]{x^2-4x+6)} is 3/4 (x^2-4x+6)^{2/3}+C