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

The integral of x^2(5x^2+x)^2 is (25x^7)/7+(5x^6)/3+(x^5)/5+C

integral of x^2(5x^2+x)^2

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

\frac{25 x ^{7}}{7} + \frac{5 x ^{6}}{3} + \frac{x ^{5}}{5} + C

Solution steps

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

Expand

x^{2} (5 x^{2} + x)^{2} : 25 x^{6} + 10 x^{5} + x^{4}

= \int 25 x^{6} + 10 x^{5} + 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 25 x^{6} d x + \int 10 x^{5} d x + \int x^{4} d x

\int 25 x^{6} d x = \frac{25 x ^{7}}{7}

\int 10 x^{5} d x = \frac{5 x ^{6}}{3}

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

= \frac{25 x ^{7}}{7} + \frac{5 x ^{6}}{3} + \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{25 x ^{7}}{7} + \frac{5 x ^{6}}{3} + \frac{x ^{5}}{5} + C

\frac{\partial}{\partial h} (\frac{1}{2} (a + b) h)

\frac{d}{d x} (\frac{2}{x ^{\frac{1}{3}}} - 2)

y^{^{''}} - 2 y^{^{'}} + 5 y = 5, y (0) = 2, y^{^{'}} (0) = 1

t an g e n t f (x) = (\frac{x - 3}{x + 1})^{2}, at x = 0

\int y e^{3 y} d y

What is the integral of x^2(5x^2+x)^2 ?
The integral of x^2(5x^2+x)^2 is (25x^7)/7+(5x^6)/3+(x^5)/5+C