What is the integral of x^2*(1+2x^3)^5 ?

The integral of x^2*(1+2x^3)^5 is 1/36 (1+2x^3)^6+C

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integral of x^2*(1+2x^3)^5

\int x^{2} \cdot (1 + 2 x^{3})^{5} d x

\frac{1}{36} (1 + 2 x^{3})^{6} + C

Solution steps

\int x^{2} (1 + 2 x^{3})^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{6} d u

= \int \frac{u ^{5}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{6} \cdot \frac{u ^{6}}{6}

Substitute back

u = 1 + 2 x^{3}

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

Simplify

\frac{1}{6} \cdot \frac{( 1 + 2 x ^{3} ) ^{6}}{6} : \frac{1}{36} (1 + 2 x^{3})^{6}

= \frac{1}{36} (1 + 2 x^{3})^{6}

Add a constant to the solution

= \frac{1}{36} (1 + 2 x^{3})^{6} + C

\frac{d}{d x} (9 x - \frac{1}{x})

f (x) = 4 x^{5} ln (x)

e^{20 t}

y^{^{'}} = \frac{( x y ^{2} )}{( 1 + x ^{2} )}

y^{^{''}} - 16 y^{^{'}} + 64 y = 0, y (0) = 1, y (1) = 0

What is the integral of x^2*(1+2x^3)^5 ?
The integral of x^2*(1+2x^3)^5 is 1/36 (1+2x^3)^6+C