What is the integral of (3x-1)/(\sqrt[3]{1-2x+3x^2)} ?

The integral of (3x-1)/(\sqrt[3]{1-2x+3x^2)} is 3/4 (1-2x+3x^2)^{2/3}+C

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integral of (3x-1)/(\sqrt[3]{1-2x+3x^2)}

\frac{3}{4} (1 - 2 x + 3 x^{2})^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

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 = 1 - 2 x + 3 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (3x-1)/(\sqrt[3]{1-2x+3x^2)} ?
The integral of (3x-1)/(\sqrt[3]{1-2x+3x^2)} is 3/4 (1-2x+3x^2)^{2/3}+C