What is the integral of x(x^4+1/4 x) ?

The integral of x(x^4+1/4 x) is (x^6)/6+(x^3)/(12)+C

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integral of x(x^4+1/4 x)

\int x (x^{4} + \frac{1}{4} x) d x

\frac{x ^{6}}{6} + \frac{x ^{3}}{12} + C

Solution steps

\int x (x^{4} + \frac{1}{4} x) d x

Expand

x (x^{4} + \frac{1}{4} x) : x^{5} + \frac{x ^{2}}{4}

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

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

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

= \frac{x ^{6}}{6} + \frac{x ^{3}}{12}

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{x ^{3}}{12} + C

\int y^{3} e^{y^{4}} d y

\frac{d}{d x} (cos ((π x)^{9}))

\frac{\partial}{\partial x} (x^{0.1})

\int (1 - x)^{2} x d x

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 7, y^{^{'}} (0) = 0

What is the integral of x(x^4+1/4 x) ?
The integral of x(x^4+1/4 x) is (x^6)/6+(x^3)/(12)+C