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

The integral of x^2(1+2x^3)^2 is (x^3)/3+(2x^6)/3+(4x^9)/9+C

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

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

\frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} + \frac{4 x ^{9}}{9} + C

Solution steps

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

Expand

x^{2} (1 + 2 x^{3})^{2} : x^{2} + 4 x^{5} + 4 x^{8}

= \int x^{2} + 4 x^{5} + 4 x^{8} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x + \int 4 x^{5} d x + \int 4 x^{8} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int 4 x^{8} d x = \frac{4 x ^{9}}{9}

= \frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} + \frac{4 x ^{9}}{9}

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{2 x ^{6}}{3} + \frac{4 x ^{9}}{9} + C

f (x) = 6 sec (x), at x = \frac{π}{3}

\frac{d}{d x} ((x^{2})^{5})

\int \frac{1 6 ^{x} + 2 0 ^{x}}{4 ^{x}} d x

x y^{^{'}} + 3 y - 3 = 0

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

What is the integral of x^2(1+2x^3)^2 ?
The integral of x^2(1+2x^3)^2 is (x^3)/3+(2x^6)/3+(4x^9)/9+C