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

The integral of (2x+3)/(\sqrt[3]{x^2+3x-1)} is (3\sqrt[3]{4})/8 (4x^2+12x-4)^{2/3}+C

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

Solution steps

Apply u-substitution

Take the constant out:

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

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} v^{\frac{2}{3}}

Substitute back

Simplify

Add a constant to the solution

\int \frac{2 x ^{3} + 8 x}{x ^{2} + 4} d x

\int (e^{t} - e^{- t})^{2} d t

\int \frac{x ^{2} - 2 x + 6}{( x ^{2} + 2 ) ( x - 2 )} d x

\int (x e^{x} - e^{x}) d x

\int \frac{x - 1}{3 x ^{2} - 6 x + 5} d x

What is the integral of (2x+3)/(\sqrt[3]{x^2+3x-1)} ?
The integral of (2x+3)/(\sqrt[3]{x^2+3x-1)} is (3\sqrt[3]{4})/8 (4x^2+12x-4)^{2/3}+C