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

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

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

\frac{3}{8} (x^{2} + 4 x + 2)^{\frac{4}{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 u-substitution

Apply the Power Rule

: \frac{3}{4} w^{\frac{4}{3}}

= \frac{1}{2} \cdot \frac{3}{4} w^{\frac{4}{3}}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{3}{4} ((x + 2)^{2} - 2)^{\frac{4}{3}} : \frac{3}{8} (x^{2} + 4 x + 2)^{\frac{4}{3}}

= \frac{3}{8} (x^{2} + 4 x + 2)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{8} (x^{2} + 4 x + 2)^{\frac{4}{3}} + C

\int \frac{( ( ( x ) + 1 ) \cdot ( x - 1 ) )}{( x )} d x

f (x) = x^{2} + 2, (- 2, 6)

\int 2^{x - 1} d x

(2) (e^{3 x x})

\int (x + 1) e^{8 x^{2} + 16 x} d x

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