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

The integral of x^2(x^2+2x)^2 is (x^7)/7+(2x^6)/3+(4x^5)/5+C

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

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

\frac{x ^{7}}{7} + \frac{2 x ^{6}}{3} + \frac{4 x ^{5}}{5} + C

Solution steps

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

Expand

x^{2} (x^{2} + 2 x)^{2} : x^{6} + 4 x^{5} + 4 x^{4}

= \int x^{6} + 4 x^{5} + 4 x^{4} 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^{6} d x + \int 4 x^{5} d x + \int 4 x^{4} d x

\int x^{6} d x = \frac{x ^{7}}{7}

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

\int 4 x^{4} d x = \frac{4 x ^{5}}{5}

= \frac{x ^{7}}{7} + \frac{2 x ^{6}}{3} + \frac{4 x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{7}}{7} + \frac{2 x ^{6}}{3} + \frac{4 x ^{5}}{5} + C

f (x) = cos (3 x) ln (x^{4})

\int \frac{3 tan ( θ ) - 4 cos ^{2} ( θ )}{cos ( θ )} d θ

x \to 0 + lim (\frac{ln ^{2} ( x )}{2})

\frac{\partial}{\partial y} (x^{2} + \frac{x}{y})

\int x^{11} x^{6} + 6 d x

What is the integral of x^2(x^2+2x)^2 ?
The integral of x^2(x^2+2x)^2 is (x^7)/7+(2x^6)/3+(4x^5)/5+C